Physics Problems with Solutions

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  • Acceleration: Tutorials with Examples

Examples with explanations on the concepts of acceleration of moving object are presented. More problems and their solutions can also be found in this website.

Average Acceleration

An object with initial velocity v 0 at time t 0 and final velocity v at time t has an average acceleration between t 0 and t given by

Examples with soltutions

What is the acceleration of an object that moves with uniform velocity? Solution: If the velocity is uniform, let us say V, then the initial and final velocities are both equal to V and the definition of the acceleration gives

A car accelerates from rest to a speed of 36 km/h in 20 seconds. What is the acceleration of the car in m/s 2 ? Solution: The initial velocity is 0 (from rest) and the final velocity is 36 km/h. Hence

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  • Velocity and Speed: Tutorials with Examples
  • Velocity and Speed: Problems with Solutions
  • Uniform Acceleration Motion: Problems with Solutions
  • Uniform Acceleration Motion: Equations with Explanations

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Acceleration Problems

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Acceleration

Practice problem 1.

  • A car is said to go "zero to sixty in six point six seconds". What is its acceleration in m/s 2 ?
  • The driver can't release his foot from the gas pedal (a.k.a. the accelerator). How many additional seconds would it take for the driver to reach 80 mph assuming the aceleration remains constant?
  • A car moving at 80 mph has a speed of 35.8 m/s. What acceleration would it have if it took 5.0 s to come to a complete stop?

Well first of all, we shouldn't be dealing with English units. They're difficult to work with, so let's convert them straight away and then do the old "plug and chug".

Since the question asked for acceleration and acceleration is a vector quantity this answer is not complete. A proper answer must include a direction as well. This is quite easy to do. Since the car is starting from rest and moving forward, its acceleration must also be forward. The ultimate, complete answer to this problem is the car is accelerating at…

a  =  4.06 m/s 2  forward

We should convert the final speed to SI units.

Use the fact that change equals rate times time, and then add that change to our velocity at the end of the previous problem. Algebra will do the rest for us.

Alternate solution. We don't need no stinkin' conversions with this method. The ratio of eighty to sixty is a simple one, namely 4 3 . From our definition of acceleration, it should be apparent that time is directly proportional to change in velocity when acceleration is constant. Thus…

This is not the answer. It is the time elapsed from the moment when the car began to move. The question was about the additional time needed, so we should subtract the time required to go from zero to sixty. Thus…

∆ t  =  8.8 s − 6.6 s  =  2.2 s

The two methods give essentially the same answer.

Quite simple. Let's do it.

Nothing surprising there except the negative sign. When a vector quantity is negative what does it mean? There are several interpretations of this, but I think mine is the best. When a vector has a negative value, it means that it points in a direction opposite that of the positive vectors. In this problem, since the positive vectors are assumed to point forward (What other direction would a normal car drive?) the acceleration must be backward. Thus the complete answer to this problem is that the car's acceleration is…

a  =  7.16 m/s 2 backward

Although it is common to assign deceleration a negative value, negative acceleration does not automatically imply deceleration. When dealing with vector quantities, any direction can be assumed positive…

up, down, right, left, forward, backward, north, south, east, west

and the corresponding opposite direction assumed negative…

down, up, left, right, backward, forward, south, north, west, east.

It won't matter which you chose as long as you are consistent throughout a problem. Don't learn any rules for assigning signs to particular directions and don't let anyone tell you that a certain direction must be positive or must be negative.

practice problem 2

Acceleration is the rate of change of velocity with time. Since velocity is a vector, this definition means acceleration is also a vector. When it comes to vectors, direction matters as much as size. In a simple one-dimensional problem like this one, directions are indicated by algebraic sign. Every quantity that points away from the batter will be positive. Every quantity that points toward him will be negative. Thus, the ball comes in at −40 m/s and goes out at +50 m/s. If we didn't pay attention to this detail, we wouldn't get the right answer.

practice problem 3

Practice problem 4.

easycalculation.com

Acceleration Examples

problem solving examples in acceleration

Acceleration is a vector quantity that measures a change in speed or direction. It is defined as a change in velocity per unit of time. Given below is the acceleration examples problems with solution for your reference to calculate acceleration in m / s 2 .

Acceleration Example Problems

Let us consider the acceleration practice problem: A car accelerates uniformly from 22.5 m/s to 46.1 m/s in 2.47 seconds. Determine the acceleration of the car.

We can calculate the Acceleration using the given formula.

Substituting the values in the formula,

Acceleration = (46.1 – 22.5 ) / 2.47 = 23.6 / 2.47 = 9.554 m/s 2

Therefore, the value of Acceleration is 9.554 m/s 2 .

Refer the below example of acceleration with solution. When an airplane accelerates down a runway at 3.20 m/s 2 to 5.41 m/s 2 for 28 s until is finally lifts off the ground, calculate its acceleration before its take off.

Substituting the values in the above given formula,

Acceleration = (5.41 – 3.20 ) / 28 = 2.21 / 28 = 0.078 m/s 2

Therefore, Acceleration is 0.078 m/s 2 .

Related Examples:

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  • Young's Modulus Calculation Examples
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problem solving examples in acceleration

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Check Your Understanding

Answer: d = 1720 m

Answer: a = 8.10 m/s/s

Answers: d = 33.1 m and v f = 25.5 m/s

Answers: a = 11.2 m/s/s and d = 79.8 m

Answer: t = 1.29 s

Answers: a = 243 m/s/s

Answer: a = 0.712 m/s/s

Answer: d = 704 m

Answer: d = 28.6 m

Answer: v i = 7.17 m/s

Answer: v i = 5.03 m/s and hang time = 1.03 s (except for in sports commericals)

Answer: a = 1.62*10 5 m/s/s

Answer: d = 48.0 m

Answer: t = 8.69 s

Answer: a = -1.08*10^6 m/s/s

Answer: d = -57.0 m (57.0 meters deep) 

Answer: v i = 47.6 m/s

Answer: a = 2.86 m/s/s and t = 30. 8 s

Answer: a = 15.8 m/s/s

Answer: v i = 94.4 mi/hr

Solutions to Above Problems

d = (0 m/s)*(32.8 s)+ 0.5*(3.20 m/s 2 )*(32.8 s) 2

Return to Problem 1

110 m = (0 m/s)*(5.21 s)+ 0.5*(a)*(5.21 s) 2

110 m = (13.57 s 2 )*a

a = (110 m)/(13.57 s 2 )

a = 8.10 m/ s 2

Return to Problem 2

d = (0 m/s)*(2.60 s)+ 0.5*(-9.8 m/s 2 )*(2.60 s) 2

d = -33.1 m (- indicates direction)

v f = v i + a*t

v f = 0 + (-9.8 m/s 2 )*(2.60 s)

v f = -25.5 m/s (- indicates direction)

Return to Problem 3

a = (46.1 m/s - 18.5 m/s)/(2.47 s)

a = 11.2 m/s 2

d = v i *t + 0.5*a*t 2

d = (18.5 m/s)*(2.47 s)+ 0.5*(11.2 m/s 2 )*(2.47 s) 2

d = 45.7 m + 34.1 m

(Note: the d can also be calculated using the equation v f 2 = v i 2 + 2*a*d)

Return to Problem 4

-1.40 m = (0 m/s)*(t)+ 0.5*(-1.67 m/s 2 )*(t) 2

-1.40 m = 0+ (-0.835 m/s 2 )*(t) 2

(-1.40 m)/(-0.835 m/s 2 ) = t 2

1.68 s 2 = t 2

Return to Problem 5

a = (444 m/s - 0 m/s)/(1.83 s)

a = 243 m/s 2

d = (0 m/s)*(1.83 s)+ 0.5*(243 m/s 2 )*(1.83 s) 2

d = 0 m + 406 m

Return to Problem 6

(7.10 m/s) 2 = (0 m/s) 2 + 2*(a)*(35.4 m)

50.4 m 2 /s 2 = (0 m/s) 2 + (70.8 m)*a

(50.4 m 2 /s 2 )/(70.8 m) = a

a = 0.712 m/s 2

Return to Problem 7

(65 m/s) 2 = (0 m/s) 2 + 2*(3 m/s 2 )*d

4225 m 2 /s 2 = (0 m/s) 2 + (6 m/s 2 )*d

(4225 m 2 /s 2 )/(6 m/s 2 ) = d

Return to Problem 8

d = (22.4 m/s + 0 m/s)/2 *2.55 s

d = (11.2 m/s)*2.55 s

Return to Problem 9

(0 m/s) 2 = v i 2 + 2*(-9.8 m/s 2 )*(2.62 m)

0 m 2 /s 2 = v i 2 - 51.35 m 2 /s 2

51.35 m 2 /s 2 = v i 2

v i = 7.17 m/s

Return to Problem 10

(0 m/s) 2 = v i 2 + 2*(-9.8 m/s 2 )*(1.29 m)

0 m 2 /s 2 = v i 2 - 25.28 m 2 /s 2

25.28 m 2 /s 2 = v i 2

v i = 5.03 m/s

To find hang time, find the time to the peak and then double it.

0 m/s = 5.03 m/s + (-9.8 m/s 2 )*t up

-5.03 m/s = (-9.8 m/s 2 )*t up

(-5.03 m/s)/(-9.8 m/s 2 ) = t up

t up = 0.513 s

hang time = 1.03 s

Return to Problem 11

(521 m/s) 2 = (0 m/s) 2 + 2*(a)*(0.840 m)

271441 m 2 /s 2 = (0 m/s) 2 + (1.68 m)*a

(271441 m 2 /s 2 )/(1.68 m) = a

a = 1.62*10 5 m /s 2

Return to Problem 12

  • (NOTE: the time required to move to the peak of the trajectory is one-half the total hang time - 3.125 s.)

First use:  v f  = v i  + a*t

0 m/s = v i  + (-9.8  m/s 2 )*(3.13 s)

0 m/s = v i  - 30.7 m/s

v i  = 30.7 m/s  (30.674 m/s)

Now use:  v f 2  = v i 2  + 2*a*d

(0 m/s) 2  = (30.7 m/s) 2  + 2*(-9.8  m/s 2 )*(d)

0 m 2 /s 2  = (940 m 2 /s 2 ) + (-19.6  m/s 2 )*d

-940  m 2 /s 2  = (-19.6  m/s 2 )*d

(-940  m 2 /s 2 )/(-19.6  m/s 2 ) = d

Return to Problem 13

-370 m = (0 m/s)*(t)+ 0.5*(-9.8 m/s 2 )*(t) 2

-370 m = 0+ (-4.9 m/s 2 )*(t) 2

(-370 m)/(-4.9 m/s 2 ) = t 2

75.5 s 2 = t 2

Return to Problem 14

(0 m/s) 2 = (367 m/s) 2 + 2*(a)*(0.0621 m)

0 m 2 /s 2 = (134689 m 2 /s 2 ) + (0.1242 m)*a

-134689 m 2 /s 2 = (0.1242 m)*a

(-134689 m 2 /s 2 )/(0.1242 m) = a

a = -1.08*10 6 m /s 2

(The - sign indicates that the bullet slowed down.)

Return to Problem 15

d = (0 m/s)*(3.41 s)+ 0.5*(-9.8 m/s 2 )*(3.41 s) 2

d = 0 m+ 0.5*(-9.8 m/s 2 )*(11.63 s 2 )

d = -57.0 m

(NOTE: the - sign indicates direction)

Return to Problem 16

(0 m/s) 2 = v i 2 + 2*(- 3.90 m/s 2 )*(290 m)

0 m 2 /s 2 = v i 2 - 2262 m 2 /s 2

2262 m 2 /s 2 = v i 2

v i = 47.6 m /s

Return to Problem 17

( 88.3 m/s) 2 = (0 m/s) 2 + 2*(a)*(1365 m)

7797 m 2 /s 2 = (0 m 2 /s 2 ) + (2730 m)*a

7797 m 2 /s 2 = (2730 m)*a

(7797 m 2 /s 2 )/(2730 m) = a

a = 2.86 m/s 2

88.3 m/s = 0 m/s + (2.86 m/s 2 )*t

(88.3 m/s)/(2.86 m/s 2 ) = t

t = 30. 8 s

Return to Problem 18

( 112 m/s) 2 = (0 m/s) 2 + 2*(a)*(398 m)

12544 m 2 /s 2 = 0 m 2 /s 2 + (796 m)*a

12544 m 2 /s 2 = (796 m)*a

(12544 m 2 /s 2 )/(796 m) = a

a = 15.8 m/s 2

Return to Problem 19

v f 2 = v i 2 + 2*a*d

(0 m/s) 2 = v i 2 + 2*(-9.8 m/s 2 )*(91.5 m)

0 m 2 /s 2 = v i 2 - 1793 m 2 /s 2

1793 m 2 /s 2 = v i 2

v i = 42.3 m/s

Now convert from m/s to mi/hr:

v i = 42.3 m/s * (2.23 mi/hr)/(1 m/s)

v i = 94.4 mi/hr

Return to Problem 20

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2.5: Motion Equations for Constant Acceleration in One Dimension

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Learning Objectives

By the end of this section, you will be able to:

  • Calculate displacement of an object that is not accelerating, given initial position and velocity.
  • Calculate final velocity of an accelerating object, given initial velocity, acceleration, and time.
  • Calculate displacement and final position of an accelerating object, given initial position, initial velocity, time, and acceleration.

We might know that the greater the acceleration of, say, a car moving away from a stop sign, the greater the displacement in a given time. But we have not developed a specific equation that relates acceleration and displacement. In this section, we develop some convenient equations for kinematic relationships, starting from the definitions of displacement, velocity, and acceleration already covered.

Four men racing up a river in their kayaks.

Notation: t , x , v , a

First, let us make some simplifications in notation. Taking the initial time to be zero, as if time is measured with a stopwatch, is a great simplification. Since elapsed time is \(\displaystyle Δt=t_f−t_0\), taking \(\displaystyle t_0=0\) means that \(\displaystyle Δt=t_f\), the final time on the stopwatch. When initial time is taken to be zero, we use the subscript 0 to denote initial values of position and velocity. That is, \(\displaystyle x_0\) is the initial position and \(\displaystyle v_0\) is the initial velocity . We put no subscripts on the final values. That is, \(\displaystyle t\) is the final time , \(\displaystyle x\) is the final position , and \(\displaystyle v\) is the final velocity . This gives a simpler expression for elapsed time—now, \(\displaystyle Δt=t\). It also simplifies the expression for displacement, which is now \(\displaystyle Δx=x−x_0\). Also, it simplifies the expression for change in velocity, which is now \(\displaystyle Δv=v−v_0\). To summarize, using the simplified notation, with the initial time taken to be zero,

\(\displaystyle Δt=t\) \(\displaystyle Δx=x−x_0\) \(\displaystyle Δv=v−v_0\)

where the subscript 0 denotes an initial value and the absence of a subscript denotes a final value in whatever motion is under consideration.

We now make the important assumption that acceleration is constant . This assumption allows us to avoid using calculus to find instantaneous acceleration. Since acceleration is constant, the average and instantaneous accelerations are equal. That is,

\[\bar{a}=a=constant,\]

so we use the symbol a for acceleration at all times. Assuming acceleration to be constant does not seriously limit the situations we can study nor degrade the accuracy of our treatment. For one thing, acceleration is constant in a great number of situations. Furthermore, in many other situations we can accurately describe motion by assuming a constant acceleration equal to the average acceleration for that motion. Finally, in motions where acceleration changes drastically, such as a car accelerating to top speed and then braking to a stop, the motion can be considered in separate parts, each of which has its own constant acceleration.

SOLVING FOR DISPLACEMENT (Δx) AND FINAL POSITION (x) FROM AVERAGE VELOCITY WHEN ACCELERATION (a) IS CONSTANT

To get our first two new equations, we start with the definition of average velocity:

\[\displaystyle \bar{v}=\frac{Δx}{Δt}\]

Substituting the simplified notation for \(\displaystyle Δx\) and \(\displaystyle Δt\) yields

Solving for \(\displaystyle x\) yields

where the average velocity is

\[\displaystyle \bar{v}=\frac{v_0+v}{2} \label{eq5}\]

with constant \(a\).

Equation \ref{eq5} reflects the fact that, when acceleration is constant, \(v\) is just the simple average of the initial and final velocities. For example, if you steadily increase your velocity (that is, with constant acceleration) from 30 to 60 km/h, then your average velocity during this steady increase is 45 km/h. Using the equation \(\displaystyle \bar{v}=\frac{v_0+v}{2}\) to check this, we see that

\(\displaystyle \bar{v}=\frac{v_0+v}{2}=\frac{30 km/h+60 km/h}{2}=45 km/h,\)

which seems logical.

Example \(\PageIndex{1}\): Calculating Displacement - How Far does the Jogger Run?

A jogger runs down a straight stretch of road with an average velocity of 4.00 m/s for 2.00 min. What is his final position, taking his initial position to be zero?

Draw a sketch.

Velocity vector arrow labeled v equals 4 point zero zero meters per second over an x axis displaying initial and final positions. Final position is labeled x equals question mark.

The final position is given by the equation

\(\displaystyle x=x_0+\bar{v}t\).

To find \(\displaystyle x\), we identify the values of \(\displaystyle x_0, \bar{v}\), and \(\displaystyle t\) from the statement of the problem and substitute them into the equation.

  • Identify the knowns. \(\displaystyle \bar{v}=4.00 m/s, Δt=2.00 min\), and \(\displaystyle x_0=0 m\).
  • Enter the known values into the equation.

Velocity and final displacement are both positive, which means they are in the same direction.

The equation \(\displaystyle x=x_0+\bar{v}t\) gives insight into the relationship between displacement, average velocity, and time. It shows, for example, that displacement is a linear function of average velocity. (By linear function, we mean that displacement depends on \(\displaystyle \bar{v}\) rather than on \(\displaystyle \bar{v}\) raised to some other power, such as \(\displaystyle \bar{v}^2\). When graphed, linear functions look like straight lines with a constant slope.) On a car trip, for example, we will get twice as far in a given time if we average 90 km/h than if we average 45 km/h.

Line graph showing displacement in meters versus average velocity in meters per second. The line is straight with a positive slope. Displacement x increases linearly with increase in average velocity v.

SOLVING FOR FINAL VELOCITY

We can derive another useful equation by manipulating the definition of acceleration.

\(\displaystyle a=\frac{Δv}{Δt}\)

Substituting the simplified notation for \(\displaystyle Δv\) and \(\displaystyle Δt\) gives us

\[\displaystyle a=\frac{v−v_0}{t}\] (constant a).

Solving for \(\displaystyle v\) yields

\[\displaystyle v=v_0+at\](constanta).

Example \(\PageIndex{2}\):Calculating Final Velocity: An Airplane Slowing Down after Landing

An airplane lands with an initial velocity of 70.0 m/s and then decelerates at \(\displaystyle 1.50 m/s^2\) for 40.0 s. What is its final velocity?

Draw a sketch. We draw the acceleration vector in the direction opposite the velocity vector because the plane is decelerating.

Velocity vector arrow pointing toward the right in the positive x direction. Initial velocity equals seventy meters per second. Final velocity equals question mark. An acceleration vector arrow pointing toward the left labeled a equals negative 1 point 50 meters per second squared.

1. Identify the knowns. \(\displaystyle v_0=70.0 m/s, a=−1.50 m/s^2, t=40.0s.\)

2. Identify the unknown. In this case, it is final velocity, \(\displaystyle v_f\).

3. Determine which equation to use. We can calculate the final velocity using the equation \(\displaystyle v=v_0+at\).

4. Plug in the known values and solve.

\(\displaystyle v=v_0+at=70.0 m/s+(−1.50 m/s^2)(40.0 s)=10.0 m/s\)

The final velocity is much less than the initial velocity, as desired when slowing down, but still positive. With jet engines, reverse thrust could be maintained long enough to stop the plane and start moving it backward. That would be indicated by a negative final velocity, which is not the case here.

An airplane moving toward the right at two points in time. At time equals 0 the velocity vector arrow points toward the right and is labeled seventy meters per second. The acceleration vector arrow points toward the left and is labeled negative 1 point 5 meters per second squared. At time equals forty seconds, the velocity arrow is shorter, points toward the right, and is labeled ten meters per second. The acceleration vector arrow is still pointing toward the left and is labeled a equals negative 1 point 5 meters per second squared.

In addition to being useful in problem solving, the equation \(\displaystyle v=v_0+at\) gives us insight into the relationships among velocity, acceleration, and time. From it we can see, for example, that

  • final velocity depends on how large the acceleration is and how long it lasts
  • if the acceleration is zero, then the final velocity equals the initial velocity (\(\displaystyle v=v_0\)), as expected (i.e., velocity is constant)
  • if a is negative, then the final velocity is less than the initial velocity

(All of these observations fit our intuition, and it is always useful to examine basic equations in light of our intuition and experiences to check that they do indeed describe nature accurately.)

MAKING CONNECTIONS: REAL-WORLD CONNECTION

Space shuttle blasting off at night.

An intercontinental ballistic missile (ICBM) has a larger average acceleration than the Space Shuttle and achieves a greater velocity in the first minute or two of flight (actual ICBM burn times are classified—short-burn-time missiles are more difficult for an enemy to destroy). But the Space Shuttle obtains a greater final velocity, so that it can orbit the earth rather than come directly back down as an ICBM does. The Space Shuttle does this by accelerating for a longer time.

SOLVING FOR FINAL POSITION WHEN VELOCITY IS NOT CONSTANT (a≠0)

We can combine the equations above to find a third equation that allows us to calculate the final position of an object experiencing constant acceleration. We start with

\(\displaystyle v=v_0+at.\)

Adding \(\displaystyle v_0\) to each side of this equation and dividing by 2 gives

\(\displaystyle \frac{v_0+v}{2}=v_0+\frac{1}{2}at\).

Since \\frac{(v_0+v}{2}=\bar{v}\) for constant acceleration, then

\(\displaystyle \bar{v}=v_0+\frac{1}{2}at\).

Now we substitute this expression for \(\displaystyle \bar{v}\) into the equation for displacement, \(\displaystyle x=x_0+\bar{v}t\), yielding

\(\displaystyle x=x_0+v_0t+\frac{1}{2}at^2\)(constant a).

Example \(\PageIndex{3}\): Calculating Displacement of an Accelerating Object - Dragsters

Dragsters can achieve average accelerations of \(\displaystyle 26.0 m/s^2\). Suppose such a dragster accelerates from rest at this rate for 5.56 s. How far does it travel in this time?

Dragster accelerating down a race track.

We are asked to find displacement, which is \(\displaystyle x\) if we take \(\displaystyle x_0\) to be zero. (Think about it like the starting line of a race. It can be anywhere, but we call it 0 and measure all other positions relative to it.) We can use the equation \(\displaystyle x=x_0+v_0t+\frac{1}{2}at^2\) once we identify \(\displaystyle v_0, a,\) and \(\displaystyle t\) from the statement of the problem.

1. Identify the knowns. Starting from rest means that \(\displaystyle v_0=0, a\) is given as \(\displaystyle 26.0m/s^2\) and t is given as 5.56 s.

2. Plug the known values into the equation to solve for the unknown x:

\(\displaystyle x=x_0+v_0t+\frac{1}{2}at^2\).

Since the initial position and velocity are both zero, this simplifies to

\(\displaystyle x=\frac{1}{2}at^2\).

Substituting the identified values of a and t gives

\(\displaystyle x=\frac{1}{2}(26.0 m/s^2)(5.56 s)^2\),

\(\displaystyle x=402 m.\)

If we convert 402 m to miles, we find that the distance covered is very close to one quarter of a mile, the standard distance for drag racing. So the answer is reasonable. This is an impressive displacement in only 5.56 s, but top-notch dragsters can do a quarter mile in even less time than this.

What else can we learn by examining the equation \(\displaystyle x=x_0+v_0t+\frac{1}{2}at^2\)? We see that:

  • displacement depends on the square of the elapsed time when acceleration is not zero. In Example, the dragster covers only one fourth of the total distance in the first half of the elapsed time
  • if acceleration is zero, then the initial velocity equals average velocity (\(\displaystyle v_0=\bar{v}\)) and \(\displaystyle x=x_0+v_0t+\frac{1}{2}at^2\) becomes \(\displaystyle x=x_0+v_0t\)

SOLVING FOR FINAL VELOCITY WHEN VELOCITY IS NOT CONSTANT (a≠0)

A fourth useful equation can be obtained from another algebraic manipulation of previous equations.

If we solve \(\displaystyle v=v_0+at\) for \(\displaystyle t\), we get

\(\displaystyle t=\frac{v−v_0}{a}\).

Substituting this and \(\displaystyle \bar{v}=\frac{v_0+v}{2}\) into \(\displaystyle x=x_0+\bar{v}t\), we get

\(\displaystyle v^2=v^2_0+2a(x−x_0)\)(constant a).

Example \(\PageIndex{4}\): Calculating Final Velocity: Dragsters

Calculate the final velocity of the dragster in Example \(\PageIndex{3}\) without using information about time.

Acceleration vector arrow pointing toward the right, labeled twenty-six point zero meters per second squared. Initial velocity equals 0. Final velocity equals question mark.

The equation \(\displaystyle v^2=v^2_0+2a(x−x_0)\) is ideally suited to this task because it relates velocities, acceleration, and displacement, and no time information is required.

1. Identify the known values. We know that \(\displaystyle v_0=0\), since the dragster starts from rest. Then we note that \(\displaystyle x−x_0=402 m\) (this was the answer in Example). Finally, the average acceleration was given to be \(\displaystyle a=26.0 m/s^2\).

2. Plug the knowns into the equation \(\displaystyle v^2=v^2_0+2a(x−x_0)\) and solve for \(\displaystyle v\).

\(\displaystyle v^2=0+2(26.0 m/s^2)(402 m).\)

\(\displaystyle v^2=2.09×10^4m^2/s^2.\)

To get \(\displaystyle v\), we take the square root:

\(\displaystyle v=\sqrt{2.09×10^4m^2/s^2}=145 m/s\).

145 m/s is about 522 km/h or about 324 mi/h, but even this breakneck speed is short of the record for the quarter mile. Also, note that a square root has two values; we took the positive value to indicate a velocity in the same direction as the acceleration.

An examination of the equation \(\displaystyle v^2=v^2_0+2a(x−x_0)\) can produce further insights into the general relationships among physical quantities:

  • The final velocity depends on how large the acceleration is and the distance over which it acts
  • For a fixed deceleration, a car that is going twice as fast doesn’t simply stop in twice the distance—it takes much further to stop. (This is why we have reduced speed zones near schools.)

Putting Equations Together

In the following examples, we further explore one-dimensional motion, but in situations requiring slightly more algebraic manipulation. The examples also give insight into problem-solving techniques. The box below provides easy reference to the equations needed.

SUMMARY OF KINEMATIC EQUATIONS (CONSTANT a)

\(\displaystyle x=x_0+\bar{v}t\)

\(\displaystyle \bar{v}=\frac{v_0+v}{2}\)

\(\displaystyle v=v_0+at\)

\(\displaystyle x=x_0+v_0t+\frac{1}{2}at^2\)

\(\displaystyle v^2=v^2_0+2a(x−x_0)\)

Example \(\PageIndex{5}\):Calculating Displacement: How Far Does a Car Go When Coming to a Halt?

On dry concrete, a car can decelerate at a rate of \(7.00 m/s^2\), whereas on wet concrete it can decelerate at only \(5.00 m/s^2\). Find the distances necessary to stop a car moving at 30.0 m/s (about 110 km/h)

  • on dry concrete and
  • on wet concrete.
  • Repeat both calculations, finding the displacement from the point where the driver sees a traffic light turn red, taking into account his reaction time of 0.500 s to get his foot on the brake.

Initial velocity equals thirty meters per second. Final velocity equals 0. Acceleration dry equals negative 7 point zero zero meters per second squared. Acceleration wet equals negative 5 point zero zero meters per second squared.

In order to determine which equations are best to use, we need to list all of the known values and identify exactly what we need to solve for. We shall do this explicitly in the next several examples, using tables to set them off.

Solution for (a)

1. Identify the knowns and what we want to solve for. We know that \(\displaystyle v_0=30.0 m/s; v=0; a=−7.00m/s^2\) (\(\displaystyle a\) is negative because it is in a direction opposite to velocity). We take \(\displaystyle x_0\) to be \(\displaystyle 0\). We are looking for displacement \(\displaystyle Δx\), or \(\displaystyle x−x_0\).

2. Identify the equation that will help up solve the problem. The best equation to use is

\(\displaystyle v^2=v^2_0+2a(x−x_0)\).

This equation is best because it includes only one unknown, \(\displaystyle x\). We know the values of all the other variables in this equation. (There are other equations that would allow us to solve for \(\displaystyle x\), but they require us to know the stopping time, \(\displaystyle t\), which we do not know. We could use them but it would entail additional calculations.)

3. Rearrange the equation to solve for \(\displaystyle x\).

\(\displaystyle x−x_0=\frac{v^2−v^2_0}{2a}\)

4. Enter known values.

\(\displaystyle x−0=\frac{0^2−(30.0 m/s)^2}{2(−7.00 m/s^2)}\)

\(\displaystyle x=64.3 m\) on dry concrete.

Solution for (b)

This part can be solved in exactly the same manner as Part A. The only difference is that the deceleration is \(\displaystyle –5.00 m/s^2\). The result is

\(\displaystyle x_{wet}=90.0 m\) on wet concrete.

Solution for (c)

Once the driver reacts, the stopping distance is the same as it is in Parts A and B for dry and wet concrete. So to answer this question, we need to calculate how far the car travels during the reaction time, and then add that to the stopping time. It is reasonable to assume that the velocity remains constant during the driver’s reaction time.

1. Identify the knowns and what we want to solve for. We know that \(\displaystyle \bar{v}=30.0 m/s; t_{reaction}=0.500s; a_{reaction}=0\). We take \(\displaystyle x_{0−reaction}\) to be 0. We are looking for \(\displaystyle x_{reaction}\).

2. Identify the best equation to use.

\(\displaystyle x=x_0+\bar{v}t\) works well because the only unknown value is \(\displaystyle x\), which is what we want to solve for.

3. Plug in the knowns to solve the equation.

\(\displaystyle x=0+(30.0 m/s)(0.500 s)=15.0 m.\)

This means the car travels 15.0 m while the driver reacts, making the total displacements in the two cases of dry and wet concrete 15.0 m greater than if he reacted instantly.

4. Add the displacement during the reaction time to the displacement when braking.

\(\displaystyle x_{braking}+x_{reaction}=x_{total}\)

a. 64.3 m + 15.0 m = 79.3 m when dry b. 90.0 m + 15.0 m = 105 m when wet

Diagram showing the various braking distances necessary for stopping a car. With no reaction time considered, braking distance is 64 point 3 meters on a dry surface and 90 meters on a wet surface. With reaction time of 0 point 500 seconds, braking distance is 79 point 3 meters on a dry surface and 105 meters on a wet surface.

The displacements found in this example seem reasonable for stopping a fast-moving car. It should take longer to stop a car on wet rather than dry pavement. It is interesting that reaction time adds significantly to the displacements. But more important is the general approach to solving problems. We identify the knowns and the quantities to be determined and then find an appropriate equation. There is often more than one way to solve a problem. The various parts of this example can in fact be solved by other methods, but the solutions presented above are the shortest.

Example \(\PageIndex{5}\): Calculating Time - A Car Merges into Traffic

Suppose a car merges into freeway traffic on a 200-m-long ramp. If its initial velocity is 10.0 m/s and it accelerates at \(\displaystyle . 00 m/s^ 2\) , how long does it take to travel the 200 m up the ramp? (Such information might be useful to a traffic engineer.)

A line segment with ends labeled x subs zero equals zero and x = two hundred. Above the line segment, the equation t equals question mark indicates that time is unknown. Three vectors, all pointing in the direction of x equals 200, represent the other knowns and unknowns. They are labeled v sub zero equals ten point zero meters per second, v equals question mark, and a equals two point zero zero meters per second squared.

1. Identify the knowns and what we want to solve for. We know that \(\displaystyle v_0=10 m/s; a=2.00 m/s^2\); and \(\displaystyle x=200 m.\)

2. We need to solve for t. Choose the best equation. \(\displaystyle x=x_0+v_0t+\frac{1}{2}at^2\) works best because the only unknown in the equation is the variable t for which we need to solve.

3. We will need to rearrange the equation to solve for \(t\). In this case, it will be easier to plug in the knowns first.

\(\displaystyle 200 m=0 m+(10.0 m/s)t+\frac{1}{2}(2.00 m/s^2)t^2\)

4. Simplify the equation. The units of meters (m) cancel because they are in each term. We can get the units of seconds (s) to cancel by taking \(\displaystyle t=ts\), where \(\displaystyle t\) is the magnitude of time and s is the unit. Doing so leaves

\(\displaystyle 200=10t+t^2.\)

5. Use the quadratic formula to solve for \(\displaystyle t\).

(a) Rearrange the equation to get 0 on one side of the equation.

\(\displaystyle t^2+10t−200=0\)

This is a quadratic equation of the form

\(\displaystyle at^2+bt+c=0,\)

where the constants are \(\displaystyle a=1.00,b=10.0\),and \(\displaystyle c=−200.\)

(b) Its solutions are given by the quadratic formula:

\(\displaystyle t=\frac{−b±\sqrt{b^2−4ac}}{2a}\).

This yields two solutions for \(\displaystyle t\), which are

\(\displaystyle t=10.0\) and \(\displaystyle −20.0\).

In this case, then, the time is \(\displaystyle t=t\) in seconds, or

\(\displaystyle t=10.0s\) and \(\displaystyle −20.0s\).

A negative value for time is unreasonable, since it would mean that the event happened 20 s before the motion began. We can discard that solution. Thus,

\(\displaystyle t=10.0s.\)

Whenever an equation contains an unknown squared, there will be two solutions. In some problems both solutions are meaningful, but in others, such as the above, only one solution is reasonable. The 10.0 s answer seems reasonable for a typical freeway on-ramp.

With the basics of kinematics established, we can go on to many other interesting examples and applications. In the process of developing kinematics, we have also glimpsed a general approach to problem solving that produces both correct answers and insights into physical relationships. Problem-Solving Basics discusses problem-solving basics and outlines an approach that will help you succeed in this invaluable task.

MAKING CONNECTIONS: TAKE-HOME EXPERIMENT—BREAKING NEWS

We have been using SI units of meters per second squared to describe some examples of acceleration or deceleration of cars, runners, and trains. To achieve a better feel for these numbers, one can measure the braking deceleration of a car doing a slow (and safe) stop. Recall that, for average acceleration, \(\displaystyle \bar{a}=Δv/Δt\). While traveling in a car, slowly apply the brakes as you come up to a stop sign. Have a passenger note the initial speed in miles per hour and the time taken (in seconds) to stop. From this, calculate the deceleration in miles per hour per second. Convert this to meters per second squared and compare with other decelerations mentioned in this chapter. Calculate the distance traveled in braking.

Exercise \(\PageIndex{1}\)

A manned rocket accelerates at a rate of \(\displaystyle 20 m/s^2\) during launch. How long does it take the rocket to reach a velocity of 400 m/s?

To answer this, choose an equation that allows you to solve for time \(\displaystyle t\), given only \(\displaystyle a, v_0\), and \(\displaystyle v\).

Rearrange to solve for \(\displaystyle t\).

\(\displaystyle t=\frac{v−v_0}{a}=\frac{400 m/s−0 m/s}{20 m/s^2}=20 s\)

  • To simplify calculations we take acceleration to be constant, so that \(\displaystyle \bar{a}=a\) at all times.
  • We also take initial time to be zero.
  • Initial position and velocity are given a subscript 0; final values have no subscript. Thus,

\(\displaystyle Δt=t\)

\(\displaystyle Δx=x−x_0\)

\(\displaystyle Δv=v−v_0\)

  • The following kinematic equations for motion with constant a are useful:
  • In vertical motion, \(\displaystyle y\) is substituted for \(\displaystyle x\).
  • 3.4 Motion with Constant Acceleration
  • Introduction
  • 1.1 The Scope and Scale of Physics
  • 1.2 Units and Standards
  • 1.3 Unit Conversion
  • 1.4 Dimensional Analysis
  • 1.5 Estimates and Fermi Calculations
  • 1.6 Significant Figures
  • 1.7 Solving Problems in Physics
  • Key Equations
  • Conceptual Questions
  • Additional Problems
  • Challenge Problems
  • 2.1 Scalars and Vectors
  • 2.2 Coordinate Systems and Components of a Vector
  • 2.3 Algebra of Vectors
  • 2.4 Products of Vectors
  • 3.1 Position, Displacement, and Average Velocity
  • 3.2 Instantaneous Velocity and Speed
  • 3.3 Average and Instantaneous Acceleration
  • 3.5 Free Fall
  • 3.6 Finding Velocity and Displacement from Acceleration
  • 4.1 Displacement and Velocity Vectors
  • 4.2 Acceleration Vector
  • 4.3 Projectile Motion
  • 4.4 Uniform Circular Motion
  • 4.5 Relative Motion in One and Two Dimensions
  • 5.2 Newton's First Law
  • 5.3 Newton's Second Law
  • 5.4 Mass and Weight
  • 5.5 Newton’s Third Law
  • 5.6 Common Forces
  • 5.7 Drawing Free-Body Diagrams
  • 6.1 Solving Problems with Newton’s Laws
  • 6.2 Friction
  • 6.3 Centripetal Force
  • 6.4 Drag Force and Terminal Speed
  • 7.2 Kinetic Energy
  • 7.3 Work-Energy Theorem
  • 8.1 Potential Energy of a System
  • 8.2 Conservative and Non-Conservative Forces
  • 8.3 Conservation of Energy
  • 8.4 Potential Energy Diagrams and Stability
  • 8.5 Sources of Energy
  • 9.1 Linear Momentum
  • 9.2 Impulse and Collisions
  • 9.3 Conservation of Linear Momentum
  • 9.4 Types of Collisions
  • 9.5 Collisions in Multiple Dimensions
  • 9.6 Center of Mass
  • 9.7 Rocket Propulsion
  • 10.1 Rotational Variables
  • 10.2 Rotation with Constant Angular Acceleration
  • 10.3 Relating Angular and Translational Quantities
  • 10.4 Moment of Inertia and Rotational Kinetic Energy
  • 10.5 Calculating Moments of Inertia
  • 10.6 Torque
  • 10.7 Newton’s Second Law for Rotation
  • 10.8 Work and Power for Rotational Motion
  • 11.1 Rolling Motion
  • 11.2 Angular Momentum
  • 11.3 Conservation of Angular Momentum
  • 11.4 Precession of a Gyroscope
  • 12.1 Conditions for Static Equilibrium
  • 12.2 Examples of Static Equilibrium
  • 12.3 Stress, Strain, and Elastic Modulus
  • 12.4 Elasticity and Plasticity
  • 13.1 Newton's Law of Universal Gravitation
  • 13.2 Gravitation Near Earth's Surface
  • 13.3 Gravitational Potential Energy and Total Energy
  • 13.4 Satellite Orbits and Energy
  • 13.5 Kepler's Laws of Planetary Motion
  • 13.6 Tidal Forces
  • 13.7 Einstein's Theory of Gravity
  • 14.1 Fluids, Density, and Pressure
  • 14.2 Measuring Pressure
  • 14.3 Pascal's Principle and Hydraulics
  • 14.4 Archimedes’ Principle and Buoyancy
  • 14.5 Fluid Dynamics
  • 14.6 Bernoulli’s Equation
  • 14.7 Viscosity and Turbulence
  • 15.1 Simple Harmonic Motion
  • 15.2 Energy in Simple Harmonic Motion
  • 15.3 Comparing Simple Harmonic Motion and Circular Motion
  • 15.4 Pendulums
  • 15.5 Damped Oscillations
  • 15.6 Forced Oscillations
  • 16.1 Traveling Waves
  • 16.2 Mathematics of Waves
  • 16.3 Wave Speed on a Stretched String
  • 16.4 Energy and Power of a Wave
  • 16.5 Interference of Waves
  • 16.6 Standing Waves and Resonance
  • 17.1 Sound Waves
  • 17.2 Speed of Sound
  • 17.3 Sound Intensity
  • 17.4 Normal Modes of a Standing Sound Wave
  • 17.5 Sources of Musical Sound
  • 17.7 The Doppler Effect
  • 17.8 Shock Waves
  • B | Conversion Factors
  • C | Fundamental Constants
  • D | Astronomical Data
  • E | Mathematical Formulas
  • F | Chemistry
  • G | The Greek Alphabet

Learning Objectives

By the end of this section, you will be able to:

  • Identify which equations of motion are to be used to solve for unknowns.
  • Use appropriate equations of motion to solve a two-body pursuit problem.

You might guess that the greater the acceleration of, say, a car moving away from a stop sign, the greater the car’s displacement in a given time. But, we have not developed a specific equation that relates acceleration and displacement. In this section, we look at some convenient equations for kinematic relationships, starting from the definitions of displacement, velocity, and acceleration. We first investigate a single object in motion, called single-body motion. Then we investigate the motion of two objects, called two-body pursuit problems .

First, let us make some simplifications in notation. Taking the initial time to be zero, as if time is measured with a stopwatch, is a great simplification. Since elapsed time is Δ t = t f − t 0 Δ t = t f − t 0 , taking t 0 = 0 t 0 = 0 means that Δ t = t f Δ t = t f , the final time on the stopwatch. When initial time is taken to be zero, we use the subscript 0 to denote initial values of position and velocity. That is, x 0 x 0 is the initial position and v 0 v 0 is the initial velocity . We put no subscripts on the final values. That is, t is the final time , x is the final position , and v is the final velocity . This gives a simpler expression for elapsed time, Δ t = t Δ t = t . It also simplifies the expression for x displacement, which is now Δ x = x − x 0 Δ x = x − x 0 . Also, it simplifies the expression for change in velocity, which is now Δ v = v − v 0 Δ v = v − v 0 . To summarize, using the simplified notation, with the initial time taken to be zero,

where the subscript 0 denotes an initial value and the absence of a subscript denotes a final value in whatever motion is under consideration.

We now make the important assumption that acceleration is constant . This assumption allows us to avoid using calculus to find instantaneous acceleration. Since acceleration is constant, the average and instantaneous accelerations are equal—that is,

Thus, we can use the symbol a for acceleration at all times. Assuming acceleration to be constant does not seriously limit the situations we can study nor does it degrade the accuracy of our treatment. For one thing, acceleration is constant in a great number of situations. Furthermore, in many other situations we can describe motion accurately by assuming a constant acceleration equal to the average acceleration for that motion. Lastly, for motion during which acceleration changes drastically, such as a car accelerating to top speed and then braking to a stop, motion can be considered in separate parts, each of which has its own constant acceleration.

Displacement and Position from Velocity

To get our first two equations, we start with the definition of average velocity:

Substituting the simplified notation for Δ x Δ x and Δ t Δ t yields

Solving for x gives us

where the average velocity is

The equation v – = v 0 + v 2 v – = v 0 + v 2 reflects the fact that when acceleration is constant, v – v – is just the simple average of the initial and final velocities. Figure 3.18 illustrates this concept graphically. In part (a) of the figure, acceleration is constant, with velocity increasing at a constant rate. The average velocity during the 1-h interval from 40 km/h to 80 km/h is 60 km/h:

In part (b), acceleration is not constant. During the 1-h interval, velocity is closer to 80 km/h than 40 km/h. Thus, the average velocity is greater than in part (a).

Solving for Final Velocity from Acceleration and Time

We can derive another useful equation by manipulating the definition of acceleration:

Substituting the simplified notation for Δ v Δ v and Δ t Δ t gives us

Solving for v yields

Example 3.7

Calculating final velocity.

Second, we identify the unknown; in this case, it is final velocity v f v f .

Last, we determine which equation to use. To do this we figure out which kinematic equation gives the unknown in terms of the knowns. We calculate the final velocity using Equation 3.12 , v = v 0 + a t v = v 0 + a t .

Figure 3.19 is a sketch that shows the acceleration and velocity vectors.

Significance

In addition to being useful in problem solving, the equation v = v 0 + a t v = v 0 + a t gives us insight into the relationships among velocity, acceleration, and time. We can see, for example, that

  • Final velocity depends on how large the acceleration is and how long it lasts
  • If the acceleration is zero, then the final velocity equals the initial velocity ( v = v 0 ), as expected (in other words, velocity is constant)
  • If a is negative, then the final velocity is less than the initial velocity

All these observations fit our intuition. Note that it is always useful to examine basic equations in light of our intuition and experience to check that they do indeed describe nature accurately.

Solving for Final Position with Constant Acceleration

We can combine the previous equations to find a third equation that allows us to calculate the final position of an object experiencing constant acceleration. We start with

Adding v 0 v 0 to each side of this equation and dividing by 2 gives

Since v 0 + v 2 = v – v 0 + v 2 = v – for constant acceleration, we have

Now we substitute this expression for v – v – into the equation for displacement, x = x 0 + v – t x = x 0 + v – t , yielding

Example 3.8

Calculating displacement of an accelerating object.

Second, we substitute the known values into the equation to solve for the unknown:

Since the initial position and velocity are both zero, this equation simplifies to

Substituting the identified values of a and t gives

What else can we learn by examining the equation x = x 0 + v 0 t + 1 2 a t 2 ? x = x 0 + v 0 t + 1 2 a t 2 ? We can see the following relationships:

  • Displacement depends on the square of the elapsed time when acceleration is not zero. In Example 3.8 , the dragster covers only one-fourth of the total distance in the first half of the elapsed time.
  • If acceleration is zero, then initial velocity equals average velocity ( v 0 = v – ) ( v 0 = v – ) , and x = x 0 + v 0 t + 1 2 a t 2 becomes x = x 0 + v 0 t . x = x 0 + v 0 t + 1 2 a t 2 becomes x = x 0 + v 0 t .

Solving for Final Velocity from Distance and Acceleration

A fourth useful equation can be obtained from another algebraic manipulation of previous equations. If we solve v = v 0 + a t v = v 0 + a t for t , we get

Substituting this and v – = v 0 + v 2 v – = v 0 + v 2 into x = x 0 + v – t x = x 0 + v – t , we get

Example 3.9

Second, we substitute the knowns into the equation v 2 = v 0 2 + 2 a ( x − x 0 ) v 2 = v 0 2 + 2 a ( x − x 0 ) and solve for v :

An examination of the equation v 2 = v 0 2 + 2 a ( x − x 0 ) v 2 = v 0 2 + 2 a ( x − x 0 ) can produce additional insights into the general relationships among physical quantities:

  • The final velocity depends on how large the acceleration is and the distance over which it acts.
  • For a fixed acceleration, a car that is going twice as fast doesn’t simply stop in twice the distance. It takes much farther to stop. (This is why we have reduced speed zones near schools.)

Putting Equations Together

In the following examples, we continue to explore one-dimensional motion, but in situations requiring slightly more algebraic manipulation. The examples also give insight into problem-solving techniques. The note that follows is provided for easy reference to the equations needed. Be aware that these equations are not independent. In many situations we have two unknowns and need two equations from the set to solve for the unknowns. We need as many equations as there are unknowns to solve a given situation.

Summary of Kinematic Equations (constant a )

Before we get into the examples, let’s look at some of the equations more closely to see the behavior of acceleration at extreme values. Rearranging Equation 3.12 , we have

From this we see that, for a finite time, if the difference between the initial and final velocities is small, the acceleration is small, approaching zero in the limit that the initial and final velocities are equal. On the contrary, in the limit t → 0 t → 0 for a finite difference between the initial and final velocities, acceleration becomes infinite.

Similarly, rearranging Equation 3.14 , we can express acceleration in terms of velocities and displacement:

Thus, for a finite difference between the initial and final velocities acceleration becomes infinite in the limit the displacement approaches zero. Acceleration approaches zero in the limit the difference in initial and final velocities approaches zero for a finite displacement.

Example 3.10

How far does a car go.

  • First, we need to identify the knowns and what we want to solve for. We know that v 0 = 30.0 m/s, v = 0, and a = −7.00 m/s 2 ( a is negative because it is in a direction opposite to velocity). We take x 0 to be zero. We are looking for displacement Δ x Δ x , or x − x 0 . Second, we identify the equation that will help us solve the problem. The best equation to use is v 2 = v 0 2 + 2 a ( x − x 0 ) . v 2 = v 0 2 + 2 a ( x − x 0 ) . This equation is best because it includes only one unknown, x . We know the values of all the other variables in this equation. (Other equations would allow us to solve for x , but they require us to know the stopping time, t , which we do not know. We could use them, but it would entail additional calculations.) Third, we rearrange the equation to solve for x : x − x 0 = v 2 − v 0 2 2 a x − x 0 = v 2 − v 0 2 2 a and substitute the known values: x − 0 = 0 2 − ( 30.0 m/s ) 2 2 ( −7.00 m/s 2 ) . x − 0 = 0 2 − ( 30.0 m/s ) 2 2 ( −7.00 m/s 2 ) . Thus, x = 64.3 m on dry concrete . x = 64.3 m on dry concrete .
  • This part can be solved in exactly the same manner as (a). The only difference is that the acceleration is −5.00 m/s 2 . The result is x wet = 90.0 m on wet concrete. x wet = 90.0 m on wet concrete.
  • When the driver reacts, the stopping distance is the same as it is in (a) and (b) for dry and wet concrete. So, to answer this question, we need to calculate how far the car travels during the reaction time, and then add that to the stopping time. It is reasonable to assume the velocity remains constant during the driver’s reaction time. To do this, we, again, identify the knowns and what we want to solve for. We know that v – = 30.0 m/s v – = 30.0 m/s , t reaction = 0.500 s t reaction = 0.500 s , and a reaction = 0 a reaction = 0 . We take x 0-reaction x 0-reaction to be zero. We are looking for x reaction x reaction . Second, as before, we identify the best equation to use. In this case, x = x 0 + v – t x = x 0 + v – t works well because the only unknown value is x , which is what we want to solve for. Third, we substitute the knowns to solve the equation: x = 0 + ( 3 0.0 m/s ) ( 0.500 s ) = 15 .0 m . x = 0 + ( 3 0.0 m/s ) ( 0.500 s ) = 15 .0 m . This means the car travels 15.0 m while the driver reacts, making the total displacements in the two cases of dry and wet concrete 15.0 m greater than if he reacted instantly. Last, we then add the displacement during the reaction time to the displacement when braking ( Figure 3.23 ), x braking + x reaction = x total , x braking + x reaction = x total , and find (a) to be 64.3 m + 15.0 m = 79.3 m when dry and (b) to be 90.0 m + 15.0 m = 105 m when wet.

Example 3.11

Calculating time.

We need to solve for t . The equation x = x 0 + v 0 t + 1 2 a t 2 x = x 0 + v 0 t + 1 2 a t 2 works best because the only unknown in the equation is the variable t , for which we need to solve. From this insight we see that when we input the knowns into the equation, we end up with a quadratic equation.

We need to rearrange the equation to solve for t , then substituting the knowns into the equation:

We then simplify the equation. The units of meters cancel because they are in each term. We can get the units of seconds to cancel by taking t = t s, where t is the magnitude of time and s is the unit. Doing so leaves

We then use the quadratic formula to solve for t ,

which yields two solutions: t = 10.0 and t = −20.0. A negative value for time is unreasonable, since it would mean the event happened 20 s before the motion began. We can discard that solution. Thus,

Check Your Understanding 3.5

A rocket accelerates at a rate of 20 m/s 2 during launch. How long does it take the rocket to reach a velocity of 400 m/s?

Example 3.12

Acceleration of a spaceship.

Then we substitute v 0 v 0 into v = v 0 + a t v = v 0 + a t to solve for the final velocity:

With the basics of kinematics established, we can go on to many other interesting examples and applications. In the process of developing kinematics, we have also glimpsed a general approach to problem solving that produces both correct answers and insights into physical relationships. The next level of complexity in our kinematics problems involves the motion of two interrelated bodies, called two-body pursuit problems .

Two-Body Pursuit Problems

Up until this point we have looked at examples of motion involving a single body. Even for the problem with two cars and the stopping distances on wet and dry roads, we divided this problem into two separate problems to find the answers. In a two-body pursuit problem , the motions of the objects are coupled—meaning, the unknown we seek depends on the motion of both objects. To solve these problems we write the equations of motion for each object and then solve them simultaneously to find the unknown. This is illustrated in Figure 3.25 .

The time and distance required for car 1 to catch car 2 depends on the initial distance car 1 is from car 2 as well as the velocities of both cars and the acceleration of car 1. The kinematic equations describing the motion of both cars must be solved to find these unknowns.

Consider the following example.

Example 3.13

Cheetah catching a gazelle.

  • Equation for the gazelle: The gazelle has a constant velocity, which is its average velocity, since it is not accelerating. Therefore, we use Equation 3.10 with x 0 = 0 x 0 = 0 : x = x 0 + v – t = v – t . x = x 0 + v – t = v – t . Equation for the cheetah: The cheetah is accelerating from rest, so we use Equation 3.13 with x 0 = 0 x 0 = 0 and v 0 = 0 v 0 = 0 : x = x 0 + v 0 t + 1 2 a t 2 = 1 2 a t 2 . x = x 0 + v 0 t + 1 2 a t 2 = 1 2 a t 2 . Now we have an equation of motion for each animal with a common parameter, which can be eliminated to find the solution. In this case, we solve for t : x = v – t = 1 2 a t 2 t = 2 v – a . x = v – t = 1 2 a t 2 t = 2 v – a . The gazelle has a constant velocity of 10 m/s, which is its average velocity. The acceleration of the cheetah is 4 m/s 2 . Evaluating t , the time for the cheetah to reach the gazelle, we have t = 2 v – a = 2 ( 10 m/s ) 4 m/s 2 = 5 s . t = 2 v – a = 2 ( 10 m/s ) 4 m/s 2 = 5 s .
  • To get the displacement, we use either the equation of motion for the cheetah or the gazelle, since they should both give the same answer. Displacement of the cheetah: x = 1 2 a t 2 = 1 2 ( 4 m/s 2 ) ( 5 ) 2 = 50 m . x = 1 2 a t 2 = 1 2 ( 4 m/s 2 ) ( 5 ) 2 = 50 m . Displacement of the gazelle: x = v – t = 10 m/s ( 5 ) = 50 m . x = v – t = 10 m/s ( 5 ) = 50 m . We see that both displacements are equal, as expected.

Check Your Understanding 3.6

A bicycle has a constant velocity of 10 m/s. A person starts from rest and begins to run to catch up to the bicycle in 30 s when the bicycle is at the same position as the person. What is the acceleration of the person?

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  • Authors: William Moebs, Samuel J. Ling, Jeff Sanny
  • Publisher/website: OpenStax
  • Book title: University Physics Volume 1
  • Publication date: Sep 19, 2016
  • Location: Houston, Texas
  • Book URL: https://openstax.org/books/university-physics-volume-1/pages/1-introduction
  • Section URL: https://openstax.org/books/university-physics-volume-1/pages/3-4-motion-with-constant-acceleration

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Solved Speed, Velocity, and Acceleration Problems

Simple problems on speed, velocity, and acceleration with descriptive answers are presented for the AP Physics 1 exam and college students. In each solution, you can find a brief tutorial. 

Speed and velocity Problems: 

Problem (1): What is the speed of a rocket that travels $8000\,{\rm m}$ in $13\,{\rm s}$?

Solution : Speed is defined in physics  as the total distance divided by the elapsed time,  so the rocket's speed is \[\text{speed}=\frac{8000}{13}=615.38\,{\rm m/s}\]

Problem (2): How long will it take if you travel $400\,{\rm km}$ with an average speed of $100\,{\rm m/s}$?

Solution : Average speed is the ratio of the total distance to the total time. Thus, the elapsed time is \begin{align*} t&=\frac{\text{total distance}}{\text{average speed}}\\ \\ &=\frac{400\times 10^{3}\,{\rm m}}{100\,{\rm m/s}}\\ \\ &=4000\,{\rm s}\end{align*} To convert it to hours, it must be divided by $3600\,{\rm s}$ which gives $t=1.11\,{\rm h}$.

Problem (3): A person walks $100\,{\rm m}$ in $5$ minutes, then $200\,{\rm m}$ in $7$ minutes, and finally $50\,{\rm m}$ in $4$ minutes. Find its average speed. 

Solution : First find its total distance traveled ($D$) by summing all distances in each section, which gets $D=100+200+50=350\,{\rm m}$. Now, by definition of average speed, divide it by the total time elapsed $T=5+7+4=16$ minutes.

But keep in mind that since the distance is in SI units, so the time traveled must also be in SI units, which is $\rm s$. Therefore, we have\begin{align*}\text{average speed}&=\frac{\text{total distance} }{\text{total time} }\\ \\ &=\frac{350\,{\rm m}}{16\times 60\,{\rm s}}\\ \\&=0.36\,{\rm m/s}\end{align*}

Problem (4): A person walks $750\,{\rm m}$ due north, then $250\,{\rm m}$ due east. If the entire walk takes $12$ minutes, find the person's average velocity. 

Solution : Average velocity , $\bar{v}=\frac{\Delta x}{\Delta t}$, is displacement divided by the elapsed time. Displacement is also a vector that obeys the addition vector rules. Thus, in this velocity problem, add each displacement to get the total displacement . 

In the first part, displacement is $\Delta x_1=750\,\hat{j}$ (due north) and in the second part $\Delta x_2=250\,\hat{i}$ (due east). The total displacement vector is $\Delta x=\Delta x_1+\Delta x_2=750\,\hat{i}+250\,\hat{j}$ with magnitude of  \begin{align*}|\Delta x|&=\sqrt{(750)^{2}+(250)^{2}}\\ \\&=790.5\,{\rm m}\end{align*} In addition, the total elapsed time is $t=12\times 60$ seconds. Therefore, the magnitude of the average velocity is \[\bar{v}=\frac{790.5}{12\times 60}=1.09\,{\rm m/s}\]

Problem (5): An object moves along a straight line. First, it travels at a velocity of $12\,{\rm m/s}$ for $5\,{\rm s}$ and then continues in the same direction with $20\,{\rm m/s}$ for $3\,{\rm s}$. What is its average speed?

Solution: Average velocity is displacement divided by elapsed time, i.e., $\bar{v}\equiv \frac{\Delta x_{tot}}{\Delta t_{tot}}$.

Here, the object goes through two stages with two different displacements, so add them to find the total displacement. Thus,\[\bar{v}=\frac{x_1 + x_2}{t_1 +t_2}\] Again, to find the displacement, we use the same equation as the average velocity formula, i.e., $x=vt$. Thus, displacements are obtained as $x_1=v_1\,t_1=12\times 5=60\,{\rm m}$ and $x_2=v_2\,t_2=20\times 3=60\,{\rm m}$. Therefore, we have \begin{align*} \bar{v}&=\frac{x_1+x_2}{t_1+t_2}\\ \\&=\frac{60+60}{5+3}\\ \\&=\boxed{15\,{\rm m/s}}\end{align*}

Problem (6): A plane flies the distance between two cities in $1$ hour and $30$ minutes with a velocity of $900\,{\rm km/h}$. Another plane covers that distance at $600\,{\rm km/h}$. What is the flight time of the second plane?

Solution: first find the distance between two cities using the average velocity formula $\bar{v}=\frac{\Delta x}{\Delta t}$ as below \begin{align*} x&=vt\\&=900\times 1.5\\&=1350\,{\rm km}\end{align*} where we wrote one hour and a half minutes as $1.5\,\rm h$. Now use again the same kinematic equation above to find the time required for another plane \begin{align*} t&=\frac xv\\ \\ &=\frac{1350\,\rm km}{600\,\rm km/h}\\ \\&=2.25\,{\rm h}\end{align*} Thus, the time for the second plane is $2$ hours and $0.25$ of an hour, which converts to minutes as $2$ hours and ($0.25\times 60=15$) minutes.

Problem (7): To reach a park located south of his jogging path, Henry runs along a 15-kilometer route. If he completes the journey in 1.5 hours, determine his speed and velocity.

Solution:  Henry travels his route to the park without changing direction along a straight line. Therefore, the total distance traveled in one direction equals the displacement, i.e, \[\text{distance traveled}=\Delta x=15\,\rm km\]Velocity is displacement divided by the time of travel \begin{align*} \text{velocity}&=\frac{\text{displacement}}{\text{time of travel}} \\\\ &=\frac{15\,\rm km}{1.5\,\rm h} \\\\ &=\boxed{10\,\rm km/h}\end{align*} and by definition, its average speed is \begin{align*} \text{speed}&=\frac{\text{distance covered}}{\text{time interval}}\\\\&=\frac{15\,\rm km}{1.5\,\rm h}\\\\&=\boxed{10\,\rm km/h}\end{align*} Thus, Henry's velocity is $10\,\rm km/h$ to the south, and its speed is $10\,\rm km/h$. As you can see, speed is simply a positive number, with units but velocity specifies the direction in which the object is moving. 

Problem (8): In 15 seconds, a football player covers the distance from his team's goal line to the opposing team's goal line and back to the midway point of the field having 100-yard-length. Find, (a) his average speed, and (b) the magnitude of the average velocity.

Solution:  The total length of the football field is $100$ yards or in meters, $L=91.44\,\rm m$. Going from one goal's line to the other and back to the midpoint of the field takes $15\,\rm s$ and covers a distance of $D=100+50=150\,\rm yd$. 

average speed and velocity at football field

Distance divided by the time of travel gets the average speed, \[\text{speed}=\frac{150\times 0.91}{15}=9.1\,\rm m/s\] To find the average velocity, we must find the displacement of the player between the initial and final points. 

The initial point is her own goal line and her final position is the midpoint of the field, so she has displaced a distance of $\Delta x=50\,\rm yd$ or $\Delta x=50\times 0.91=45.5\,\rm m$. Therefore, her velocity is calculated as follows \begin{align*} \text{velocity}&=\frac{\text{displacement}}{\text{time elapsed}} \\\\ &=\frac{45.5\,\rm m}{15\,\rm s} \\\\&=\boxed{3.03\quad \rm m/s}\end{align*} Contrary to the previous problem, here the motion is not in one direction, hence, the displacement is not equal to the distance traveled. Accordingly, the average speed is not equal to the magnitude of the average velocity.

Problem (9): You begin at a pillar and run towards the east (the positive $x$ direction) for $250\,\rm m$ at an average speed of $5\,\rm m/s$. After that, you run towards the west for $300\,\rm m$ at an average speed of $4\,\rm m/s$ until you reach a post. Calculate (a) your average speed from pillar to post, and (b) your average velocity from pillar to post. 

Solution : First, you traveled a distance of $L_1=250\,\rm m$ toward east (or $+x$ direction) at $5\,\rm m/s$. Time of travel in this route is obtained as follows \begin{align*} t_1&=\frac{L_1}{v_1}\\\\ &=\frac{250}{5}\\\\&=50\,\rm s\end{align*} Likewise, traveling a distance of $L_2=300\,\rm m$ at $v_2=4\,\rm m/s$ takes \[t_2=\frac{300}{4}=75\,\rm s\]  (a) Average speed is defined as the distance traveled (or path length) divided by the total time of travel \begin{align*} v&=\frac{\text{path length}}{\text{time of travel}} \\\\ &=\frac{L_1+L_2}{t_1+t_2}\\\\&=\frac{250+300}{50+75} \\\\&=4.4\,\rm m/s\end{align*} Therefore, you travel between these two pillars in $125\,\rm s$ and with an average speed of $4.4\,\rm m/s$. 

(b) Average velocity requires finding the displacement between those two points. In the first case, you move $250\,\rm m$ toward $+x$ direction, i.e., $L_1=+250\,\rm m$. Similarly, on the way back, you move $300\,\rm m$ toward the west ($-x$ direction) or $L_2=-300\,\rm m$. Adding these two gives us the total displacement between the initial point and the final point, \begin{align*} L&=L_1+L_2 \\\\&=(+250)+(-300) \\\\ &=-50\,\rm m\end{align*} The minus sign indicates that you are generally displaced toward the west. 

Finally, the average velocity is obtained as follows: \begin{align*} \text{average velocity}&=\frac{\text{displacement}}{\text{time of travel}} \\\\ &=\frac{-50}{125} \\\\&=-0.4\,\rm m/s\end{align*} A negative average velocity indicating motion to the left along the $x$-axis. 

This speed problem better makes it clear to us the difference between average speed and average speed. Unlike average speed, which is always a positive number, the average velocity in a straight line can be either positive or negative. 

Problem (10): What is the average speed for the round trip of a car moving uphill at 40 km/h and then back downhill at 60 km/h? 

Solution : Assuming the length of the hill to be $L$, the total distance traveled during this round trip is $2L$ since $L_{up}=L_{down}=L$. However, the time taken for going uphill and downhill was not provided. We can write them in terms of the hill's length $L$ as $t=\frac L v$. 

Applying the definition of average speed gives us \begin{align*} v&=\frac{\text{distance traveled}}{\text{total time}} \\\\ &=\frac{L_{up}+L_{down}}{t_{up}+t_{down}} \\\\ &=\cfrac{2L}{\cfrac{L}{v_{up}}+\cfrac{L}{v_{down}}} \end{align*} By reorganizing this expression, we obtain a formula that is useful for solving similar problems in the AP Physics 1 exams. \[\text{average speed}=\frac{2v_{up} \times v_{down}}{v_{up}+v_{down}}\] Substituting the numerical values into this, yields \begin{align*} v&=\frac{2(40\times 60)}{40+60} \\\\ &=\boxed{48\,\rm m/s}\end{align*} What if we were asked for the average velocity instead? During this round trip, the car returns to its original position, and thus its displacement, which defines the average velocity, is zero. Therefore, \[\text{average velocity}=0\,\rm m/s\]

Acceleration Problems

Problem (9): A car moves from rest to a speed of $45\,\rm m/s$ in a time interval of $15\,\rm s$. At what rate does the car accelerate? 

Solution : The car is initially at rest, $v_1=0$, and finally reaches $v_2=45\,\rm m/s$ in a time interval $\Delta t=15\,\rm s$. Average acceleration is the change in velocity, $\Delta v=v_2-v_1$, divided by the elapsed time $\Delta t$, so \[\bar{a}=\frac{45-0}{15}=\boxed{3\,\rm m/s^2} \] 

Problem (10): A car moving at a velocity of $15\,{\rm m/s}$, uniformly slows down. It comes to a complete stop in $10\,{\rm s}$. What is its acceleration?

Solution:  Let the car's uniform velocity be $v_1$ and its final velocity $v_2=0$.   Average acceleration is the difference in velocities divided by the time taken, so we have: \begin{align*}\bar{a}&=\frac{\Delta v}{\Delta t}\\\\&=\frac{v_2-v_1}{\Delta t}\\\\&=\frac{0-15}{10}\\\\ &=\boxed{-1.5\,{\rm m/s^2}}\end{align*}The minus sign indicates the direction of the acceleration vector, which is toward the $-x$ direction.

Problem (11): A car moves from rest to a speed of $72\,{\rm km/h}$ in $4\,{\rm s}$. Find the acceleration of the car.

Solution: Known: $v_1=0$, $v_2=72\,{\rm km/h}$, $\Delta t=4\,{\rm s}$.  Average acceleration is defined as the difference in velocities divided by the time interval between those points \begin{align*}\bar{a}&=\frac{v_2-v_1}{t_2-t_1}\\\\&=\frac{20-0}{4}\\\\&=5\,{\rm m/s^2}\end{align*} In above, we converted $\rm km/h$ to the SI unit of velocity ($\rm m/s$) as \[1\,\frac{km}{h}=\frac {1000\,m}{3600\,s}=\frac{10}{36}\, \rm m/s\] so we get \[72\,\rm km/h=72\times \frac{10}{36}=20\,\rm m/s\] 

Problem (12): A race car accelerates from an initial velocity of $v_i=10\,{\rm m/s}$ to a final velocity of $v_f = 30\,{\rm m/s}$ in a time interval of $2\,{\rm s}$. Determine its average acceleration.

Solution:  A change in the velocity of an object $\Delta v$ over a time interval $\Delta t$ is defined as an average acceleration. Known: $v_i=10\,{\rm m/s}$, $v_f = 30\,{\rm m/s}$, $\Delta t=2\,{\rm s}$. Applying definition of average acceleration, we get \begin{align*}\bar{a}&=\frac{v_f-v_i}{\Delta t}\\&=\frac{30-10}{2}\\&=10\,{\rm m/s^2}\end{align*}

Problem (13): A motorcycle starts its trip along a straight line with a velocity of $10\,{\rm m/s}$ and ends with $20\,{\rm m/s}$ in the opposite direction in a time interval of $2\,{\rm s}$. What is the average acceleration of the car?

Solution:  Known: $v_i=10\,{\rm m/s}$, $v_f=-20\,{\rm m/s}$, $\Delta t=2\,{\rm s}$, $\bar{a}=?$. Using average acceleration definition we have \begin{align*}\bar{a}&=\frac{v_f-v_i}{\Delta t}\\\\&=\frac{(-20)-10}{2}\\\\ &=\boxed{-15\,{\rm m/s^2}}\end{align*}Recall that in the definition above, velocities are vector quantities. The final velocity is in the opposite direction from the initial velocity so a negative must be included.

Problem (14): A ball is thrown vertically up into the air by a boy. After $4$ seconds, it reaches the highest point of its path. How fast does the ball leave the boy's hand?

Solution : At the highest point, the ball has zero speed, $v_2=0$. It takes the ball $4\,\rm s$ to reach that point. In this problem, our unknown is the initial speed of the ball, $v_1=?$. Here, the ball accelerates at a constant rate of $g=-9.8\,\rm m/s^2$ in the presence of gravity.

When the ball is tossed upward, the only external force that acts on it is the gravity force. 

Using the average acceleration formula $\bar{a}=\frac{\Delta v}{\Delta t}$ and substituting the numerical values into this, we will have \begin{gather*} \bar{a}=\frac{\Delta v}{\Delta t} \\\\ -9.8=\frac{0-v_1}{4} \\\\ \Rightarrow \boxed{v_1=39.2\,\rm m/s} \end{gather*} Note that $\Delta v=v_2-v_1$. 

Problem (15): A child drops crumpled paper from a window. The paper hit the ground in $3\,\rm s$. What is the velocity of the crumpled paper just before it strikes the ground? 

Solution : The crumpled paper is initially in the child's hand, so $v_1=0$. Let its speed just before striking be $v_2$. In this case, we have an object accelerating down in the presence of gravitational force at a constant rate of $g=-9.8\,\rm m/s^2$. Using the definition of average acceleration, we can find $v_2$ as below \begin{gather*} \bar{a}=\frac{\Delta v}{\Delta t} \\\\ -9.8=\frac{v_2-0}{3} \\\\ \Rightarrow v_2=3\times (-9.8)=\boxed{-29.4\,\rm m/s} \end{gather*} The negative shows us that the velocity must be downward, as expected!

Problem (16): A car travels along the $x$-axis for $4\,{\rm s}$ at an average velocity of $10\,{\rm m/s}$ and $2\,{\rm s}$ with an average velocity of $30\,{\rm m/s}$ and finally $4\,{\rm s}$ with an average velocity $25\,{\rm m/s}$. What is its average velocity across the whole path?

Solution: There are three different parts with different average velocities. Assume each trip is done in one dimension without changing direction. Thus, displacements associated with each segment are the same as the distance traveled in that direction and is calculated as below: \begin{align*}\Delta x_1&=v_1\,\Delta t_1\\&=10\times 4=40\,{\rm m}\\ \\ \Delta x_2&=v_2\,\Delta t_2\\&=30\times 2=60\,{\rm m}\\ \\ \Delta x_3&=v_3\,\Delta t_3\\&=25\times 4=100\,{\rm m}\end{align*}Now use the definition of average velocity, $\bar{v}=\frac{\Delta x_{tot}}{\Delta t_{tot}}$, to find it over the whole path\begin{align*}\bar{v}&=\frac{\Delta x_{tot}}{\Delta t_{tot}}\\ \\&=\frac{\Delta x_1+\Delta x_2+\Delta x_3}{\Delta t_1+\Delta t_2+\Delta t_3}\\ \\&=\frac{40+60+100}{4+2+4}\\ \\ &=\boxed{20\,{\rm m/s}}\end{align*}

Problem (17): An object moving along a straight-line path. It travels with an average velocity $2\,{\rm m/s}$ for $20\,{\rm s}$ and $12\,{\rm m/s}$ for $t$ seconds. If the total average velocity across the whole path is $10\,{\rm m/s}$, then find the unknown time $t$.

Solution: In this velocity problem, the whole path $\Delta x$ is divided into two parts $\Delta x_1$ and $\Delta x_2$ with different average velocities and times elapsed, so the total average velocity across the whole path is obtained as \begin{align*}\bar{v}&=\frac{\Delta x}{\Delta t}\\\\&=\frac{\Delta x_1+\Delta x_2}{\Delta t_1+\Delta t_2}\\\\&=\frac{\bar{v}_1\,t_1+\bar{v}_2\,t_2}{t_1+t_2}\\\\10&=\frac{2\times 20+12\times t}{20+t}\\\Rightarrow t&=80\,{\rm s}\end{align*}

Note : whenever a moving object, covers distances $x_1,x_2,x_3,\cdots$ in $t_1,t_2,t_3,\cdots$ with constant or average velocities $v_1,v_2,v_3,\cdots$ along a straight-line without changing its direction, then its total average velocity across the whole path is obtained by one of the following formulas

  • Distances and times are known:\[\bar{v}=\frac{x_1+x_2+x_3+\cdots}{t_1+t_2+t_3+\cdots}\]
  • Velocities and times are known: \[\bar{v}=\frac{v_1\,t_1+v_2\,t_2+v_3\,t_3+\cdots}{t_1+t_2+t_3+\cdots}\]
  • Distances and velocities are known:\[\bar{v}=\frac{x_1+x_2+x_3+\cdots}{\frac{x_1}{v_1}+\frac{x_2}{v_2}+\frac{x_3}{v_3}+\cdots}\]

Problem (18): A car travels one-fourth of its path with a constant velocity of $10\,{\rm m/s}$, and the remaining with a constant velocity of $v_2$. If the total average velocity across the whole path is $16\,{\rm m/s}$, then find the $v_2$?

Solution: This is the third case of the preceding note. Let the length of the path be $L$ so \begin{align*}\bar{v}&=\frac{x_1+x_2}{\frac{x_1}{v_1}+\frac{x_2}{v_2}}\\\\16&=\frac{\frac 14\,L+\frac 34\,L}{\frac{\frac 14\,L}{10}+\frac{\frac 34\,L}{v_2}}\\\\\Rightarrow v_2&=20\,{\rm m/s}\end{align*}

Problem (19): An object moves along a straight-line path. It travels for $t_1$ seconds with an average velocity $50\,{\rm m/s}$ and $t_2$ seconds with a constant velocity of $25\,{\rm m/s}$. If the total average velocity across the whole path is $30\,{\rm m/s}$, then find the ratio $\frac{t_2}{t_1}$?

Solution: the velocities and times are known, so we have \begin{align*}\bar{v}&=\frac{v_1\,t_1+v_2\,t_2}{t_1+t_2}\\\\30&=\frac{50\,t_1+25\,t_2}{t_1+t_2}\\\\ \Rightarrow \frac{t_2}{t_1}&=4\end{align*} 

Read more related articles:  

Kinematics Equations: Problems and Solutions

Position vs. Time Graphs

Velocity vs. Time Graphs

In the following section, some sample AP Physics 1 problems on acceleration are provided.

Problem (20): An object moves with constant acceleration along a straight line. If its velocity at instant of $t_1 = 3\,{\rm s}$ is $10\,{\rm m/s}$ and at the moment of $t_2 = 8\,{\rm s}$ is $20\,{\rm m/s}$, then what is its initial speed?

Solution: Let the initial speed at time $t=0$ be $v_0$. Now apply average acceleration definition in the time intervals $[t_0,t_1]$ and $[t_0,t_2]$ and equate them.\begin{align*}\text{average acceleration}\ \bar{a}&=\frac{\Delta v}{\Delta t}\\\\\frac{v_1 - v_0}{t_1-t_0}&=\frac{v_2-v_0}{t_2-t_0}\\\\ \frac{10-v_0}{3-0}&=\frac{20-v_0}{8-0}\\\\ \Rightarrow v_0 &=4\,{\rm m/s}\end{align*} In the above, $v_1$ and $v_2$ are the velocities at moments $t_1$ and $t_2$, respectively. 

Problem (21): For $10\,{\rm s}$, the velocity of a car that travels with a constant acceleration, changes from $10\,{\rm m/s}$ to $30\,{\rm m/s}$. How far does the car travel?

Solution: Known: $\Delta t=10\,{\rm s}$, $v_1=10\,{\rm m/s}$ and $v_2=30\,{\rm m/s}$. 

Method (I) Without computing the acceleration: Recall that in the case of constant acceleration, we have the following kinematic equations for average velocity and displacement:\begin{align*}\text{average velocity}:\,\bar{v}&=\frac{v_1+v_2}{2}\\\text{displacement}:\,\Delta x&=\frac{v_1+v_2}{2}\times \Delta t\\\end{align*}where $v_1$ and $v_2$ are the velocities in a given time interval. Now we have \begin{align*} \Delta x&=\frac{v_1+v_2}{2}\\&=\frac{10+30}{2}\times 10\\&=200\,{\rm m}\end{align*}

Method (II) with computing acceleration: Using the definition of average acceleration, first determine it as below \begin{align*}\bar{a}&=\frac{\Delta v}{\Delta t}\\\\&=\frac{30-10}{10}\\\\&=2\,{\rm m/s^2}\end{align*} Since the velocities at the initial and final points of the problem are given so use the below time-independent kinematic equation to find the required displacement \begin{align*} v_2^{2}-v_1^{2}&=2\,a\Delta x\\\\ (30)^{2}-(10)^{2}&=2(2)\,\Delta x\\\\ \Rightarrow \Delta x&=\boxed{200\,{\rm m}}\end{align*}

Problem (22): A car travels along a straight line with uniform acceleration. If its velocity at the instant of $t_1=2\,{\rm s}$ is $36\,{\rm km/s}$ and at the moment $t_2=6\,{\rm s}$ is $72\,{\rm km/h}$, then find its initial velocity (at $t_0=0$)?

Solution: Use the equality of definition of average acceleration $a=\frac{v_f-v_i}{t_f-t_i}$ in the time intervals $[t_0,t_1]$ and $[t_0,t_2]$ to find the initial velocity as below \begin{align*}\frac{v_2-v_0}{t_2-t_0}&=\frac{v_1-v_0}{t_1-t_0}\\\\ \frac{20-v_0}{6-0}&=\frac{10-v_0}{2-0}\\\\ \Rightarrow v_0&=\boxed{5\,{\rm m/s}}\end{align*}

All these kinematic problems on speed, velocity, and acceleration are easily solved by choosing an appropriate kinematic equation. Keep in mind that these motion problems in one dimension are of the uniform or constant acceleration type. Projectiles are also another type of motion in two dimensions with constant acceleration.

Author:   Dr. Ali Nemati

Date Published: 9/6/2020

Updated: Jun 28,  2023

© 2015 All rights reserved. by Physexams.com

  • Physics Formulas

Acceleration Formula

One may have perceived that pushing a terminally ill bus can give it a sudden start. That’s because lift provides an upward push when it starts. Here velocity changes and this is acceleration! Henceforth, the frame accelerates. Acceleration is described as the rate of change of velocity of an object. A body’s acceleration is the final result of all the forces being applied to the body, as defined by Newton’s second law. Acceleration is a vector quantity that is described as the frequency at which a body’s velocity changes.

Formula of Acceleration

Acceleration is the rate of change in velocity to the change in time. It is denoted by symbol a and is articulated as-

acceleration formula 1

The  S.I  unit for acceleration is meter per second square or m/s 2 .

velocity in terms of acceleration

  • Final Velocity is v
  • Initial velocity is u
  • Acceleration is a
  • Time taken is t
  • Distance traveled is s

Acceleration Solved Examples

Underneath we have provided some sample numerical based on acceleration which might aid you to get an idea of how the formula is made use of:

Problem 1:  A toy car accelerates from 3 m/s to 5 m/s in 5 s. What is its acceleration? Answer:

Given: Initial Velocity u = 3  m/s, Final Velocity v = 5m/s, Time taken t = 5s.

Acceleration formula 5

Problem 2:  A stone is released into the river from a bridge. It takes 4s for the stone to touch the river’s water surface. Compute the height of the bridge from the water level.

(Initial Velocity) u = 0 (because the stone was at rest), t = 4s (t is Time taken) a = g = 9.8 m/s 2 , (a is Acceleration due to gravity) distance traveled by stone = Height of bridge  = s The distance covered is articulated by

Acceleration formula 9

s = 0 + 1/2 × 9.8 × 4 = 19.6 m/s 2

Therefore, s = 19.6 m/s 2

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Mathematics LibreTexts

3.1: Velocity and Acceleration

  • Last updated
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  • Page ID 89728

  • Joel Feldman, Andrew Rechnitzer and Elyse Yeager
  • University of British Columbia

If you are moving along the \(x\)–axis and your position at time \(t\) is \(x(t)\text{,}\) then your velocity at time \(t\) is \(v(t)=x'(t)\) and your acceleration at time \(t\) is \(a(t)=v'(t) = x''(t)\text{.}\)

Example 3.1.1 Velocity as derivative of position.

Suppose that you are moving along the \(x\)–axis and that at time \(t\) your position is given by

\begin{align*} x(t)&=t^3-3t+2. \end{align*}

We're going to try and get a good picture of what your motion is like. We can learn quite a bit just by looking at the sign of the velocity \(v(t)=x'(t)\) at each time \(t\text{.}\)

  • If \(x'(t) \gt 0\text{,}\) then at that instant \(x\) is increasing, i.e. you are moving to the right.
  • If \(x'(t)=0\text{,}\) then at that instant you are not moving at all.
  • If \(x'(t) \lt 0\text{,}\) then at that instant \(x\) is decreasing, i.e. you are moving to the left.

From the given formula for \(x(t)\) it is straight forward to work out the velocity

\begin{align*} v(t) = x'(t) &=3t^2-3=3(t^2-1)=3(t+1)(t-1) \end{align*}

This is zero only when \(t=-1\) and when \(t=+1\text{;}\) at no other value 1  of \(t\) can this polynomial be equal zero. Consequently in any time interval that does not include either \(t=-1\) or \(t=+1\text{,}\) \(v(t)\) takes only a single sign 2 . So

  • For all \(t \lt -1\text{,}\) both \((t+1)\) and \((t-1)\) are negative (sub in, for example, \(t=-10\)) so the product \(v(t)=x'(t)=3(t+1)(t-1) \gt 0\text{.}\)
  • For all \(-1 \lt t \lt 1\text{,}\) the factor \((t+1) \gt 0\) and the factor \((t-1) \lt 0\) (sub in, for example, \(t=0\)) so the product \(v(t)=x'(t)=3(t+1)(t-1) \lt 0\text{.}\)
  • For all \(t \gt 1\text{,}\) both \((t+1)\) and \((t-1)\) are positive (sub in, for example, \(t=+10\)) so the product \(v(t)=x'(t)=3(t+1)(t-1) \gt 0\text{.}\)

The figure below gives a summary of the sign information we have about \(t-1\text{,}\) \(t+1\) and \(x'(t)\text{.}\)

It is now easy to put together a mental image of your trajectory.

  • For \(t\) large and negative (i.e. far in the past), \(x(t)\) is large and negative and \(v(t)\) is large and positive. For example 3 , when \(t=-10^6\text{,}\) \(x(t)\approx t^3=- 10^{18}\) and \(v(t)\approx 3t^2 = 3\cdot 10^{12}\text{.}\) So you are moving quickly to the right.
  • For \(t \lt -1\text{,}\) \(v(t)=x'(t) \gt 0\) so that \(x(t)\) is increasing and you are moving to the right.
  • At \(t=-1\text{,}\) \(v(-1)=0\) and you have come to a halt at position \(x(-1)=(-1)^3-3(-1)+2=4\text{.}\)
  • For \(-1 \lt t \lt 1\text{,}\) \(v(t)=x'(t) \lt 0\) so that \(x(t)\) is decreasing and you are moving to the left.
  • At \(t=+1\text{,}\) \(v(1)=0\) and you have again come to a halt, but now at position \(x(1)=1^3-3+2=0\text{.}\)
  • For \(t \gt 1\text{,}\) \(v(t)=x'(t) \gt 0\) so that \(x(t)\) is increasing and you are again moving to the right.
  • For \(t\) large and positive (i.e. in the far future), \(x(t)\) is large and positive and \(v(t)\) is large and positive. For example 4 , when \(t=10^6\text{,}\) \(x(t)\approx t^3= 10^{18}\) and \(v(t)\approx 3t^2 = 3\cdot 10^{12}\text{.}\) So you are moving quickly to the right.

Here is a sketch of the graphs of \(x(t)\) and \(v(t)\text{.}\) The heavy lines in the graphs indicate when you are moving to the right — that is where \(v(t)=x'(t)\) is positive.

And here is a schematic picture of the whole trajectory.

Example 3.1.2 Position and velocity from acceleration.

In this example we are going to figure out how far a body falling from rest will fall in a given time period.

  • time in seconds by \(t\text{,}\)
  • mass in kilograms by \(m\text{,}\)
  • distance fallen (in metres) at time \(t\) by \(s(t)\text{,}\) velocity (in m/sec) by \(v(t)=s'(t)\) and acceleration (in m/sec\(^2\)) by \(a(t)=v'(t)=s''(t)\text{.}\)

It makes sense to choose a coordinate system so that the body starts to fall at \(t=0\text{.}\)

\begin{gather*} \text{the force applied to the body at time } t = m \cdot a(t). \end{gather*}

\begin{align*} \text{the force due to gravity acting on a body of mass } m &= m \cdot g. \end{align*}

\begin{align*} v(0) &= 0. \end{align*}

\begin{align*} m\cdot a(t) &= \text{force due to gravity}\\ m \cdot v'(t) &= m \cdot g & \text{ cancel the } m\\ v'(t) &=g \end{align*}

\begin{gather*} v(t) = gt + c \end{gather*}

for any constant \(c\text{.}\) One can then verify 6  that \(v'(t)=g\text{.}\) Using the fact that \(v(0)=0\) we must then have \(c=0\) and so

\begin{align*} v(t) &= gt. \end{align*}

\begin{align*} s'(t) &= v(t) = g \cdot t. \end{align*}

\begin{align*} s(t) &= \frac{g}{2} t^2 + c \end{align*}

\begin{align*} s(t) &= \frac{g}{2} t^2 & \text{but $g=9.8$, so}\\ &= 4.9 t^2, \end{align*}

Let's now do a similar but more complicated example.

Example 3.1.3 Stoping distance of a braking car.

A car's brakes can decelerate the car at 64000\(\textrm{km/hr}^2\text{.}\) How fast can the car be driven if it must be able to stop within a distance of 50m?

Before getting started, notice that there is a small “trick” in this problem — several quantities are stated but their units are different. The acceleration is stated in kilometres per hour\(^2\text{,}\) but the distance is stated in metres. Whenever we come across a “real world” problem 8  we should be careful of the units used.

  • time (in hours) by \(t\text{,}\)
  • the position of the car (in kilometres) at time \(t\) by \(x(t)\text{,}\) and
  • the velocity (in kilometres per hour) by is \(v(t)\text{.}\)

We can also choose a coordinate system such that \(x(0)=0\) and the car starts braking at time \(t=0\text{.}\)

  • We are told that, at maximum braking, the acceleration \(v'(t)=x''(t)\) of the car is \(-64000\text{.}\)
  • We need to determine the maximum initial velocity \(v(0)\) so that the stopping distance is at most \(50m = 0.05km\) (being careful with our units). Let us call the stopping distance \(x_{stop}\) which is really \(x(t_{stop})\) where \(t_{stop}\) is the stopping time.
  • In order to determine \(x_{stop}\) we first need to determine \(t_{stop}\text{,}\) which we will do by assuming maximum braking from a, yet to be determined, initial velocity of \(v(0)=q\) m/sec.

\begin{align*} v'(t) &= -64000 \end{align*}

This equation is very similar to the ones we had to solve in Example 3.1.2 just above.

As we did there  9  Now is a good time to go back and have a read of that example. , we are going to just guess \(v(t)\text{.}\) First, we just guess one function whose derivative is \(-64000\text{,}\) namely \(-64000 t\text{.}\) Next we observe that, since the derivative of a constant is zero, any function of the form

\begin{gather*} v(t) = -64000\,t + c \end{gather*}

with constant \(c\text{,}\) has the correct derivative. Finally, the requirement that the initial velocity \(v(0)=q\)" forces \(c=q\text{,}\) so

\begin{gather*} v(t) = q - 64000\,t \end{gather*}

\begin{align*} 0 = v(t_{stop}) &= q-64000 \cdot t_{stop} & \text{ and so}\\ t_{stop} &= \frac{q}{64000}. \end{align*}

\begin{align*} x'(t) &= v(t) = q - 64000t. \end{align*}

\begin{align*} x(t) &= qt - 32000t^2 + \text{constant} \end{align*}

\begin{align*} x(t) &= qt - 32000 t^2. \end{align*}

\begin{align*} x_{stop} &= x(t_{stop}) = q t_{stop} - 32000 t_{stop}^2\\ &= \frac{q^2}{64000} - \frac{32000 q^2}{64000^2}\\ &= \frac{q^2}{64000} \left(1 - \frac{1}{2} \right)\\ &= \frac{q^2}{2 \times 64000} \end{align*}

\begin{align*} x_{stop} = \frac{q^2}{2 \times 64000} &\leq \frac{5}{100}\\ q^2 &\leq \frac{2 \times 64000 \times 5}{100} = \frac{64000 \times 10}{100} = 6400 \end{align*}

Exercise \(\PageIndex{1}\)

Suppose you throw a ball straight up in the air, and its height from \(t=0\) to \(t=4\) is given by \(h(t)=-4.9t^2+19.6t\text{.}\) True or false: at time \(t=2\text{,}\) the acceleration of the ball is 0.

Exercise \(\PageIndex{2}\)

Suppose an object is moving with a constant acceleration. It takes ten seconds to accelerate from \(1\;\frac{\mathrm{m}}{\mathrm{s}}\) to \(2\;\frac{\mathrm{m}}{\mathrm{s}}\text{.}\) How long does it take to accelerate from \(2\;\frac{\mathrm{m}}{\mathrm{s}}\) to \(3\;\frac{\mathrm{m}}{\mathrm{s}}\text{?}\) How long does it take to accelerate from \(3\;\frac{\mathrm{m}}{\mathrm{s}}\) to \(13\;\frac{\mathrm{m}}{\mathrm{s}}\text{?}\)

Exercise \(\PageIndex{3}\)

Let \(s(t)\) be the position of a particle at time \(t\text{.}\) True or false: if \(s''(a) \gt 0\) for some \(a\text{,}\) then the particle's speed is increasing when \(t=a\text{.}\)

Exercise \(\PageIndex{4}\)

Let \(s(t)\) be the position of a particle at time \(t\text{.}\) True or false: if \(s'(a) \gt 0\) and \(s''(a) \gt 0\) for some \(a\text{,}\) then the particle's speed is increasing when \(t=a\text{.}\)

For this section, we will ask you a number of questions that have to do with objects falling on Earth. Unless otherwise stated, you should assume that an object falling through the air has an acceleration due to gravity of 9.8 meters per second per second.

Exercise \(\PageIndex{5}\)

A flower pot rolls out of a window 10m above the ground. How fast is it falling just as it smacks into the ground?

Exercise \(\PageIndex{6}\)

You want to know how deep a well is, so you drop a stone down and count the seconds until you hear it hit bottom.

  • If the stone took \(x\) seconds to hit bottom, how deep is the well?
  • Suppose you think you dropped the stone down the well, but actually you tossed it down, so instead of an initial velocity of 0 metres per second, you accidentally imparted an initial speed of \(1\) metres per second. What is the actual depth of the well, if the stone fell for \(x\) seconds?

Exercise \(\PageIndex{7}\)

You toss a key to your friend, standing two metres away. The keys initially move towards your friend at 2 metres per second, but slow at a rate of 0.25 metres per second per second. How much time does your friend have to react to catch the keys? That is--how long are the keys flying before they reach your friend?

Exercise \(\PageIndex{8}\)

A car is driving at 100 kph, and it brakes with a deceleration of \(50000 \frac{\mathrm{km}}{\mathrm{hr}^2}\text{.}\) How long does the car take to come to a complete stop?

Exercise \(\PageIndex{9}\)

You are driving at 120 kph, and need to stop in 100 metres. How much deceleration do your brakes need to provide? You may assume the brakes cause a constant deceleration.

Exercise \(\PageIndex{10}\)

You are driving at 100 kph, and apply the brakes steadily, so that your car decelerates at a constant rate and comes to a stop in exactly 7 seconds. What was your speed one second before you stopped?

Exercise \(\PageIndex{11}\)

About 8.5 minutes after liftoff, the US space shuttle has reached orbital velocity, 17 500 miles per hour. Assuming its acceleration was constant, how far did it travel in those 8.5 minutes?

Source: http://www.nasa.gov/mission_pages/shuttle/shuttlemissions/sts121/launch/qa-leinbach.html

Exercise \(\PageIndex{12}\)

A pitching machine has a dial to adjust the speed of the pitch. You rotate it so that it pitches the ball straight up in the air. How fast should the ball exit the machine, in order to stay in the air exactly 10 seconds?

You may assume that the ball exits from ground level, and is acted on only by gravity, which causes a constant deceleration of 9.8 metres per second.

Exercise \(\PageIndex{13}\)

A peregrine falcon can dive at a speed of 325 kph. If you were to drop a stone, how high up would you have to be so that the stone reached the same speed in its fall?

Exercise \(\PageIndex{14}\)

You shoot a cannon ball into the air with initial velocity \(v_0\text{,}\) and then gravity brings it back down (neglecting all other forces). If the cannon ball made it to a height of 100m, what was \(v_0\text{?}\)

Exercise \(\PageIndex{15}\)

Suppose you are driving at 120 kph, and you start to brake at a deceleration of \(50 000\) kph per hour. For three seconds you steadily increase your deceleration to \(60 000\) kph per hour. (That is, for three seconds, the rate of change of your deceleration is constant.) How fast are you driving at the end of those three seconds?

Exercise \(\PageIndex{16}\)

You jump up from the side of a trampoline with an initial upward velocity of \(1\) metre per second. While you are in the air, your deceleration is a constant \(9.8\) metres per second per second due to gravity. Once you hit the trampoline, as you fall your speed decreases by \(4.9\) metres per second per second. How many seconds pass between the peak of your jump and the lowest part of your fall on the trampoline?

Exercise \(\PageIndex{17}\)

Suppose an object is moving so that its velocity doubles every second. Give an expression for the acceleration of the object.

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Physics library

Course: physics library   >   unit 1.

  • Average velocity for constant acceleration
  • Acceleration of aircraft carrier take-off
  • Airbus A380 take-off distance
  • Deriving displacement as a function of time, acceleration, and initial velocity
  • Plotting projectile displacement, acceleration, and velocity
  • Projectile height given time
  • Deriving max projectile displacement given time
  • Impact velocity from given height
  • Viewing g as the value of Earth's gravitational field near the surface

What are the kinematic formulas?

  • Choosing kinematic equations
  • Setting up problems with constant acceleration
  • Kinematic formulas in one-dimension

What is a freely flying object—i.e., a projectile?

How do you select and use a kinematic formula, how do you derive the first kinematic formula, v = v 0 + a t ‍   , how do you derive the second kinematic formula, δ x = ( v + v 0 2 ) t ‍   , how do you derive the third kinematic formula, δ x = v 0 t + 1 2 a t 2 ‍   , how do you derive the fourth kinematic formula, v 2 = v 0 2 + 2 a δ x ‍   , what's confusing about the kinematic formulas, what do solved examples involving the kinematic formulas look like, example 1: first kinematic formula, v = v 0 + a t ‍  , example 2: second kinematic formula, δ x = ( v + v 0 2 ) t ‍  , example 3: third kinematic formula, δ x = v 0 t + 1 2 a t 2 ‍  , example 4: fourth kinematic formula, v 2 = v 0 2 + 2 a δ x ‍  , want to join the conversation.

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Kinematic Equations: Explanation, Review, and Examples

  • The Albert Team
  • Last Updated On: April 29, 2022

problem solving examples in acceleration

Now that you’ve learned about displacement, velocity, and acceleration, you’re well on your way to being able to describe just about any motion you could observe around you with physics. All that’s left is to learn how these values really play into each other. We know a few ways to move between them, but they’re all pretty limited. What happens if you need to find displacement, but only know acceleration and time? We don’t have a way to combine all of those values yet. Enter the four kinematic equations. 

What We Review

The Kinematic Equations

The following four kinematic equations come up throughout physics from the earliest high school class to the highest level college course:

Don’t let all of these numbers and symbols intimidate you. We’ll talk through each one – what they mean and when we use them. By the end of this post, you’ll be a master of understanding and implementing each of these physics equations. Let’s start with defining what all of those symbols mean. 

The First Kinematic Equation

This physics equation would be read as “the final velocity is equal to the initial velocity plus acceleration times time”. All it means is that if you have constant acceleration for some amount of time, you can find the final velocity. You’ll use this one whenever you’re looking at changing velocities with a constant acceleration.

The Second Kinematic Equation

This one is read as “displacement equals final velocity plus initial velocity divided by two times time”. You’ll use this one whenever you don’t have an acceleration to work with but you need to relate a changing velocity to a displacement.

The Third Kinematic Equation

This one may look a bit scarier as it is longer than the others, but it is read as “displacement equals initial velocity times time plus one half acceleration times time squared”. All it means is that our displacement can be related to our initial velocity and a constant acceleration without having to find the final velocity. You’ll use this one when final velocity is the only value you don’t know yet.

It is worth noting that this kinematic equation has another popular form: x=x_{0}+v_{0}t+\frac{1}{2}at^{2} . While that may seem even more intimidating, it’s actually exactly the same. The only difference here is that we have split up \Delta x into x-x_{0} and then solved to get x on its own. This version can be particularly helpful if you’re looking specifically for a final or initial position rather than just an overall displacement.

The Fourth Kinematic Equation

Our last kinematic equation is read as “final velocity squared equals initial velocity squared plus two times acceleration times displacement”. It’s worth noting that this is the only kinematic equation without time in it. Many starting physicists have been stumped by reaching a problem without a value for time. While staring at an equation sheet riddled with letters and numbers can be overwhelming, remembering you have this one equation without time will come up again and again throughout your physics career.

It may be worth noting that all of these are kinematic equations for constant acceleration. While this may seem like a limitation, we learned before that high school physics courses generally utilize constant acceleration so we don’t need to worry about it changing yet. If you do find yourself in a more advanced course, new physics equations will be introduced at the appropriate times.

How to Approach a Kinematics Problem

So now that we have all of these different kinematic equations, how do we know when to use them? How can we look at a physics word problem and know which of these equations to apply? You must use problem-solving steps. Follow these few steps when trying to solve any complex problems, and you won’t have a problem.

Step 1: Identify What You Know

This one probably seems obvious, but skipping it can be disastrous to any problem-solving endeavor. In physics problems, this just means pulling out values and directions. If you can add the symbol to go with the value (writing t=5\text{ s} instead of just 5\text{ s} , for example), even better. It’ll save time and make future steps even easier.

Step 2: Identify the Goal

In physics, this means figuring out what question you’re actually being asked. Does the question want you to find the displacement? The acceleration? How long did the movement take? Figure out what you’re being asked to do and then write down the symbol of the value you’re solving for with a question mark next to it ( t=\text{?} , for example). Again, this feels obvious, but it’s also a vital step.

Step 3: Gather Your Tools

Generally, this means a calculator and an equation. You’ll want to look at all of the symbols you wrote down and pick the physics equation for all of them, including the unknown value. Writing everything down beforehand will make it easier to pull a relevant equation than having to remember what values you need while searching for the right equation. You can use the latter method, but you’re far more likely to make a mistake and feel frustrated that way.

Step 4: Put it all Together

Plug your values into your equation and solve for the unknown value. This will usually be your last step, though you may find yourself having to repeat it a few times for exceptionally complex problems. That probably won’t come up for quite a while, though. After you’ve found your answer, it’s generally a good idea to circle it to make it obvious. That way, whoever is grading you can find it easily and you can easily keep track of which problems you’ve already completed while flipping through your work.

Kinematic Equation 1: Review and Examples

To learn how to solve problems with these new, longer equations, we’ll start with v=v_{0}+at . This kinematic equation shows a relationship between final velocity, initial velocity, constant acceleration, and time. We will explore this equation as it relates to physics word problems. This equation is set up to solve for velocity, but it can be rearranged to solve for any of the values it contains. For this physics equation and the ones following, we will look at one example finding the variable that has already been isolated and one where a new variable needs to be isolated using the steps we just outlined. So, let’s jump into applying this kinematic equation to a real-world problem.

A car sits at rest waiting to merge onto a highway. When they have a chance, they accelerate at 4\text{ m/s}^2 for 7\text{ s} . What is the car’s final velocity?

problem solving examples in acceleration

We have a clearly stated acceleration and time, but there’s no clearly defined initial velocity here. Instead, we have to take this from context. We know that the car “sits at rest” before it starts moving. This means that our initial velocity in this situation is zero. Other context clues for an object starting at rest is if it is “dropped” or if it “falls”. Our other known values will be even easier to pull as we were actually given numerical values. Now it’s time to put everything into a list.

  • v_{0}=0\text{ m/s}
  • a=4\text{ m/s}^2
  • t=7\text{ s}

Our goal here was clearly stated: find the final velocity. We’ll still want to list that out so we can see exactly what symbols we have to work with on this problem.

We already know which of the kinematic equations we’re using, but if we didn’t, this would be where we search our equation sheet for the right one. Regardless, we’ll want to write that down too.

Step 4: Put it All Together

At this point, we’ll plug all of our values into our kinematic equation. If you’re working on paper, there’s no need to repeat anything we’ve put above. That being said, for the purposes of digital organization and so you can see the full problem in one spot, we will be rewriting things here.

Now let’s get a bit trickier with a problem that will require us to rearrange our kinematic equation.

A ball rolls toward a hill at 3\text{ m/s} . It rolls down the hill for 5\text{ s} and has a final velocity of 18\text{ m/s} . What was the ball’s acceleration as it rolled down the hill?

Just like before, we’ll make a list of our known values:

  • v_{0}=3\text{ m/s}
  • t=5\text{ s}
  • v=18\text{ m/s}

Again, our goal was clearly stated, so let’s add it to our list:

We already know which equation we’re using, but let’s pretend we didn’t. We know that we need to solve for acceleration, but if you look at our original list of kinematic equations, there isn’t one that’s set up to solve for acceleration:

This begs the question, how to find acceleration (or any value) that hasn’t already been solved for? The answer is to rearrange an equation. First, though, we need to pick the right one. We start by getting rid of the second equation in this list as it doesn’t contain acceleration at all. Our options are now:

  • \Delta x=v_{0}t+\dfrac{1}{2}at^{2}
  • v^{2}=v_{0}^{2}+2a\Delta x

Now we’ll need to look at the first list we made of what we know. We know the initial velocity, time, and final velocity. There’s only one equation that has all the values we’re looking for and all of the values we know with none that we don’t. This is the first kinematic equation:

In this case, we knew the kinematic equation coming in so this process of elimination wasn’t necessary, but that won’t often be the case in the future. You’ll likely have to find the correct equation far more often than you’ll have it handed to you. It’s best to practice finding it now while we only have a few equations to work with.

Like before, we’ll be rewriting all of our relevant information below, but you won’t need to if you’re working on paper.

Although you can plug in values before rearranging the equation, in physics, you’ll usually see the equation be rearranged before values are added. This is mainly done to help keep units where they’re supposed to be and to avoid any mistakes that could come from moving numbers and units rather than just a variable. We’ll be taking the latter approach here. Follow the standard PEMDAS rules for rearranging the equation and then write it with the variable we’ve isolated on the left. While that last part isn’t necessary, it is a helpful organizational practice:

For a review of solving literal equations, visit this post ! Now we can plug in those known values and solve:

Kinematic Equation 2: Review and Examples

Next up in our four kinematics equations is \Delta x=\dfrac{v+v_{0}}{2} t . This one relates an object’s displacement to its average velocity and time. The right-hand side shows the final velocity plus the initial velocity divided by two – the sum of some values divided by the number of values, or the average. Although this equation doesn’t directly show a constant acceleration, it still assumes it. Applying this equation when acceleration isn’t constant can result in some error so best not to apply it if a changing acceleration is mentioned.

A car starts out moving at 10\text{ m/s} and accelerates to a velocity of 24\text{ m/s} . What displacement does the car cover during this velocity change if it occurs over 10\text{ s} ?

  • v_{0}=10\text{ m/s}
  • v=24\text{ m/s}
  • t=10\text{ s}
  • \Delta x=\text{?}
  • \Delta x=\dfrac{v+v_{0}}{2} t

This time around we won’t repeat everything here. Instead, We’ll jump straight into plugging in our values and solving our problem:

problem solving examples in acceleration

A ball slows down from 15\text{ m/s} to 3\text{ m/s} over a distance of 36\text{ m} . How long did this take?

  • v_{0}=15\text{ m/s}
  • v=3\text{ m/s}
  • \Delta x=36\text{ m}

We don’t have a kinematic equation for time specifically, but we learned before that we can rearrange certain equations to solve for different variables. So, we’ll pull the equation that has all of the values we need and isolate the variable we want later:

Again, we won’t be rewriting anything, but we will begin by rearranging our equation to solve for time:

Now we can plug in our known values and solve for time.

Kinematic Equation 3: Review and Examples

Our next kinematic equation is \Delta x=v_{0}t+\frac{1}{2}at^{2} . This time we are relating our displacement to our initial velocity, time, and acceleration. The only odd thing you may notice is that it doesn’t include our final velocity, only the initial. This equation will come in handy when you don’t have a final velocity that was stated either directly as a number or by a phrase indicating the object came to rest. Just like before, we’ll use this equation first to find a displacement, and then we’ll rearrange it to find a different value.

A rocket is cruising through space with a velocity of 50\text{ m/s} and burns some fuel to create a constant acceleration of 10\text{ m/s}^2 . How far will it have traveled after 5\text{ s} ?

  • v_{0}=50\text{ m/s}
  • a=10\text{ m/s}^2
  • \Delta x=v_{0}t+\frac{1}{2}at^{2}

At this point, it appears that these problems seem to be quite long and take several steps. While that is an inherent part of physics in many ways, it will start to seem simpler as time goes on. This problem presents the perfect example. While it may have been easy to combine lines 4 and 5 mathematically, they were shown separately here to make sure the process was as clear as possible. While you should always show all of the major steps of your problem-solving process, you may find that you are able to combine some of the smaller steps after some time of working with these kinematic equations.

Later in its journey, the rocket is moving along at 20\text{ m/s} when it has to fire its thrusters again. This time it covers a distance of 500\text{ m} in 10\text{ s} . What was the rocket’s acceleration during this thruster burn?

  • v_{0}=20\text{ m/s}
  • \Delta x=500\text{ m}

As usual, we’ll begin by rearranging the equation, this time to solve for acceleration.

Now we can plug in our known values to find the value of our acceleration.

Kinematic Equation 4: Review and Examples

The last of the kinematic equations that we will look at is v^{2}=v_{0}^{2}+2a\Delta x . This one is generally the most complicated looking, but it’s also incredibly important as it is our only kinematic equation that does not involve time. It relates final velocity, initial velocity, acceleration, and displacement without needing a time over which a given motion occurred. For this equation, as with the others, let’s solve it as is and then rearrange it to solve for a different variable.

A car exiting the highway begins with a speed of 25\text{ m/s} and travels down a 100\text{ m} long exit ramp with a deceleration (negative acceleration) of 3\text{ m/s}^2 . What is the car’s velocity at the end of the exit ramp?

  • v_{0}=25\text{ m/s}
  • \Delta x=100\text{ m}
  • a=-3\text{ m/s}^2

Note that our acceleration here is a negative value. That is because our problem statement gave us a deceleration instead of an acceleration. Whenever you have a deceleration, you’ll make the value negative to use it as an acceleration in your problem-solving. This also tells us that our final velocity should be less than our initial velocity so we can add that to the list of what we know as well.

  • Final velocity will be less than initial.

Being able to know something to help check your answer at the end is what makes this subject a bit easier than mathematics for some students.

While we generally try to not have any operations going on for the isolated variable, sometimes it’s actually easier that way. Having your isolated variable raised to a power is generally a time to solve before simplifying. This may seem like an arbitrary rule, and in some ways it is, but as you continue through your physics journey you’ll come up with your own practices for making problem-solving easier.

Now that we have both sides simplified, we’ll take the square root to eliminate the exponent on the left-hand side:

If we remember back at the beginning, we said that our final velocity would have to be less than our initial velocity because the problem statement told us that we were decelerating. Our initial velocity was 25\text{ m/s} which is, indeed, greater than 5\text{ m/s} so our answer checks out.

problem solving examples in acceleration

A ghost is sliding a wrench across a table to terrify the mortal onlooker. The wrench starts with a velocity of 2\text{ m/s} and accelerates to a velocity of 5\text{ m/s} over a distance of 7\text{ m} . What acceleration did the ghost move the wrench with?

  • v_{0}=2\text{ m/s}
  • v=5\text{ m/s}
  • \Delta x=7\text{ m}

We can also make an inference about our acceleration here – that it will be positive. Not every problem will tell you clearly the direction of the acceleration, but if your final velocity is greater than your initial velocity, you can be sure that your acceleration will be positive.

  • Positive acceleration

You’ll get better at picking up on subtle hints like this as you continue your physics journey and your brain starts naturally picking up on some patterns. You’ll likely find this skill more and more helpful as it develops and as problems get more difficult.

We’ll start by rearranging our equation to solve for acceleration.

As usual, now that we’ve rearranged our equation, we can plug in our values.

Again, we can go back to the beginning when we said our acceleration would be a positive number and confirm that it is. 

Problem-Solving Strategies

At this point, you’re likely getting the sense that physics will be a lot of complex problem-solving. If so, your senses are correct. In many ways, physics is the science of explaining nature with mathematical equations. There’s a lot that goes into developing and applying these equations, but at this point in your physics career, you’ll find that the majority of your time will likely be spent on applying equations to word problems. If you feel that your problem-solving skills could still use some honing, check out more examples and strategies from this post by the Physics Classroom or through this video-guided tutorial from Khan Academy.

That was a lot of equations and examples to take in. Eventually, whether you’re figuring out how to find a constant acceleration or how to solve velocity when you don’t have a value for time, you’ll know exactly which of the four kinematic equations to apply and how. Just keep the problem-solving steps we’ve used here in mind, and you’ll be able to get through your physics course without any unsolvable problems.

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    AboutTranscript. Acceleration (a) is the change in velocity (Δv) over the change in time (Δt), represented by the equation a = Δv/Δt. This allows you to measure how fast velocity changes in meters per second squared (m/s^2). Acceleration is also a vector quantity, so it includes both magnitude and direction. Created by Sal Khan.

  9. 3.2 Representing Acceleration with Equations and Graphs

    The second kinematic equation, another expression for average velocity v¯, v ¯, is simply the initial velocity plus the final velocity divided by two. This equation is only valid for constant acceleration. v¯ = v0 +vf 2 v ¯ = v 0 + v f 2. 3.5. Now we come to our main focus of this chapter; namely, the kinematic equations that describe ...

  10. Acceleration Examples

    Example 1: Let us consider the acceleration practice problem: A car accelerates uniformly from 22.5 m/s to 46.1 m/s in 2.47 seconds. Determine the acceleration of the car. Solution: We can calculate the Acceleration using the given formula. Formula: Acceleration = ( Final velocity - Initial velocity) / Time Substituting the values in the formula,

  11. How to Find Acceleration: Review and Examples

    Example 1: Acceleration of a Car Speeding Up Example 2: Acceleration of a Car Slowing Down Determining the Direction of Acceleration Positive Acceleration Negative Acceleration Examples: Identify the Direction of Acceleration Practice Using the Acceleration Formula in Word Problems Example 1: Acceleration of a Falling Ball

  12. Kinematic Equations: Sample Problems and Solutions

    Upton Chuck is riding the Giant Drop at Great America. If Upton free falls for 2.60 seconds, what will be his final velocity and how far will he fall? See Answer See solution below. A race car accelerates uniformly from 18.5 m/s to 46.1 m/s in 2.47 seconds. Determine the acceleration of the car and the distance traveled. See Answer

  13. 2.5: Motion Equations for Constant Acceleration in One Dimension

    In addition to being useful in problem solving, the equation \(\displaystyle v=v_0+at\) gives us insight into the relationships among velocity, acceleration, and time. ... We have been using SI units of meters per second squared to describe some examples of acceleration or deceleration of cars, runners, and trains. To achieve a better feel for ...

  14. 3.4 Motion with Constant Acceleration

    You might guess that the greater the acceleration of, say, a car moving away from a stop sign, the greater the car's displacement in a given time. But, we have not developed a specific equation that relates acceleration and displacement.

  15. Solved Speed, Velocity, and Acceleration Problems

    Solution: first find the distance between two cities using the average velocity formula \bar {v}=\frac {\Delta x} {\Delta t} vˉ = ΔtΔx as below \begin {align*} x&=vt\\&=900\times 1.5\\&=1350\, {\rm km}\end {align*} x = vt = 900 ×1.5 = 1350km where we wrote one hour and a half minutes as 1.5\,\rm h 1.5h.

  16. Acceleration Formula With Solved Examples

    Problem 1: A toy car accelerates from 3 m/s to 5 m/s in 5 s. What is its acceleration? Answer: Given: Initial Velocity u = 3 m/s, Final Velocity v = 5m/s, Time taken t = 5s. Problem 2: A stone is released into the river from a bridge. It takes 4s for the stone to touch the river's water surface.

  17. Uniform Acceleration Motion: Problems with Solutions

    Solution to Problem 1 Problem 2: With an initial velocity of 20 km/h, a car accelerated at 8 m/s 2 for 10 seconds. a) What is the position of the car at the end of the 10 seconds? b) What is the velocity of the car at the end of the 10 seconds? Solution to Problem 2 A car accelerates uniformly from 0 to 72 km/h in 11.5 seconds.

  18. Setting up problems with constant acceleration

    Setting up problems with constant acceleration. Onur drops a basketball from a height of 10 m on Mars, where the acceleration due to gravity has a magnitude of 3.7 m s 2 . We want to know how many seconds the basketball is in the air before it hits the ground. We can ignore air resistance. Which kinematic formula would be most useful to solve ...

  19. 3.1: Velocity and Acceleration

    Stage 1. Exercise 3.1.1. Suppose you throw a ball straight up in the air, and its height from t = 0 to t = 4 is given by h(t) = − 4.9t2 + 19.6t. True or false: at time t = 2, the acceleration of the ball is 0. Exercise 3.1.2. Suppose an object is moving with a constant acceleration.

  20. Acceleration

    Learn the acceleration formula and its units and see example problems solving for acceleration. Updated: 11/21/2023 Table of Contents. What is Acceleration? Acceleration Formula ...

  21. Position, velocity, acceleration problems and solutions

    Ad blocker detected Position, velocity, acceleration problems and solutions When solving a Physics problem in general and one of Kinematics in particular, it is important that you follow an order. Get used to being organized when you solve problems, and you will see how it gives good results.

  22. What are the kinematic formulas? (article)

    1. v = v 0 + a t. 2. Δ x = ( v + v 0 2) t. 3. Δ x = v 0 t + 1 2 a t 2. 4. v 2 = v 0 2 + 2 a Δ x. Since the kinematic formulas are only accurate if the acceleration is constant during the time interval considered, we have to be careful to not use them when the acceleration is changing.

  23. Kinematic Equations: Explanation, Review, and Examples

    This one probably seems obvious, but skipping it can be disastrous to any problem-solving endeavor. In physics problems, this just means pulling out values and directions. If you can add the symbol to go with the value (writing t=5\text{ s} instead of just 5\text{ s}, for example), even better. It'll save time and make future steps even easier.