problem solving in uniformly accelerated motion

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From rest, 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 1

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. a) What is the acceleration of the car in m/s 2 ? b) What is the position of the car by the time it reaches the velocity of 72 km/h? Solution to Problem 3

An object is thrown straight down from the top of a building at a speed of 20 m/s. It hits the ground with a speed of 40 m/s. a) How high is the building? b) How long was the object in the air? Solution to Problem 4

A train brakes from 40 m/s to a stop over a distance of 100 m. a) What is the acceleration of the train? b) How much time does it take the train to stop? Solution to Problem 5

A boy on a bicycle increases his velocity from 5 m/s to 20 m/s in 10 seconds. a) What is the acceleration of the bicycle? b) What distance was covered by the bicycle during the 10 seconds? Solution to Problem 6

a) How long does it take an airplane to take off if it needs to reach a speed on the ground of 350 km/h over a distance of 600 meters (assume the plane starts from rest)? b) What is the acceleration of the airplane over the 600 meters? Solution to Problem 7

Starting from a distance of 20 meters to the left of the origin and at a velocity of 10 m/s, an object accelerates to the right of the origin for 5 seconds at 4 m/s 2 . What is the position of the object at the end of the 5 seconds of acceleration? Solution to Problem 8

What is the smallest distance, in meters, needed for an airplane touching the runway with a velocity of 360 km/h and an acceleration of -10 m/s 2 to come to rest? Solution to Problem 9

Problem 10:

To approximate the height of a water well, Martha and John drop a heavy rock into the well. 8 seconds after the rock is dropped, they hear a splash caused by the impact of the rock on the water. What is the height of the well. (Speed of sound in air is 340 m/s). Solution to Problem 10

Problem 11:

A rock is thrown straight up and reaches a height of 10 m. a) How long was the rock in the air? b) What is the initial velocity of the rock? Solution to Problem 11

Problem 12:

A car accelerates from rest at 1.0 m/s 2 for 20.0 seconds along a straight road . It then moves at a constant speed for half an hour. It then decelerates uniformly to a stop in 30.0 s. Find the total distance covered by the car. Solution to Problem 12

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• Velocity and Speed: Tutorials with Examples
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6.2: Uniformly Accelerated Motion

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• Page ID 6961

• Jeremy Tatum
• University of Victoria

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Before studying motion in a resisting medium, a brief review of uniformly accelerating motion might be in order. That is, motion in which the resistance is zero. Any formulas that we develop for motion in a resisting medium must go to the formulas for uniformly accelerated motion as the resistance approaches zero.

One may imagine a situation in which a body starts with speed \( v_{0}\) and then accelerates at a rate \( a\). One may ask three questions:

How fast is it moving after time \( t\) ? How far has it moved in time \( t\) ? How fast is it moving after it has covered a distance \( x\) ?

The answers to these questions are well known to any student of physics:

\[ \ v= v_{0} +at, \tag{6.2.1}\label{eq:6.2.1} \]

\[ \ v= v_{0}t +\frac{1}{2}at^2, \tag{6.2.2}\label{eq:6.2.2} \]

\[ \ v^2= v_{0}^2 +2ax. \tag{6.2.3}\label{eq:6.2.3} \]

Since the acceleration is uniform, there is no need to use calculus to derive these. The first follows immediately from the meaning of acceleration. Distance travelled is the area under a speed : time graph. Figure VI.1 shows a speed : time graph for constant acceleration, and Equation \( \ref{eq:6.2.2}\) is obvious from a glance at the graph. Equation \( \ref{eq:6.2.3}\) can be obtained by elimination of \( t\) between Equations \( \ref{eq:6.2.1}\) and \( \ref{eq:6.2.2}\). (It can also be deduced from energy considerations, though that is rather putting the cart before the horse.)

Nevertheless, although calculus is not necessary, it is instructive to see how calculus can be used to analyse uniformly accelerated motion, since calculus will be necessary in less simple situations. We shall be using calculus to answer the three questions posed earlier in the section.

For uniformly accelerated motion, the Equation of motion is

\[ \ \ddot{x}=a. \tag{6.2.4}\label{eq:6.2.4} \]

To answer the first question, we write \( \ddot{x}\) as \( \frac{dv}{dt}\), and then the integral (with initial condition \( x=0\) when \( t=0\)) is

\[ \ v = v_{0} + at. \tag{6.2.5}\label{eq:6.2.5} \]

This is the first time integral. Next, we write \( v\) as \( \frac{dv}{dt}\) and integrate again with respect to time, to get

\[ \ x = v_{0}t + \frac{1}{2}at^2. \tag{6.2.6}\label{eq:6.2.6} \]

This is the second time integral. To obtain the answer to the third question, which will be called the space integral , we must remember to write \( \ddot{x} \) as \( v \frac{dv}{dt}\). Thus the Equation of motion (Equation \( \ref{eq:6.2.4}\)) is

\[ \ v\frac{dv}{dx}= a. \tag{6.2.7}\label{eq:6.2.7} \]

When this is integrated with respect to \( x\) (with initial condition \( v=v_{0}\) when \( x=0\)) we obtain

\[ \ v^{2} = v^2_{0} +2ax. \tag{6.2.8}\label{eq:6.2.8} \]

This is the space integral.

Here are a few quick examples of problems in uniformly accelerated motion. It is probably a good idea to work in algebra and obtain algebraic solutions to each problem. That is, even if you are told that the initial speed is 15 ms -1 , call it \(v_{0}\), or, if you are told that the height is 900 feet, call it \( h\). You will probably find it helpful to sketch graphs either of distance versus time or speed versus time in most of the problems. One last little hint: Remember that the two solutions of a quadratic Equation are equal if \( b^{2}=4ac\).

Example \(\PageIndex{1}\)

A body is dropped from rest. The last third of the distance before it hits the ground is covered in time T. Show that the time taken for the entire fall to the ground is 5.45T.

Example \(\PageIndex{2}\)

The Lady is 8 metres from the bus stop, when the Bus, starting from rest at the bus stop, starts to move off with an acceleration of 0.4 m s -2 . What is the least speed at which the Lady must run in order to catch the Bus?

Answer: 2.53ms -1 .

Example \(\PageIndex{3}\)

A parachutist is descending at a constant speed of 10 feet per second. When she is at a height of 900 feet, her friend, directly below her, throws an apple up to her. What is the least speed at which he must throw the apple in order for it to reach her? How long does it take to reach her, what height is she at then, and what is the relative speed of parachutist and apple? Assume \( g\) = 32 ft s -2 . Neglect air resistance for the apple (but not for the parachutist!)

Answer:230fts -1 , 7.5s, 825 ft, 0fts -1 .

Example \(\PageIndex{4}\)

A lunar explorer performs the following experiment on the Moon in order to determine the gravitational acceleration \( g\) there. He tosses a lunar rock upwards at an initial speed of 15 m s -1 . Eight seconds later he tosses another rock upwards at an initial speed of 10 m s -1 . He observes that the rocks collide 16.32 seconds after the launch of the first rock. Calculate g and also the height of the collision.

Answer: 1.64ms -2 , 26.4m

Example \(\PageIndex{5}\)

Mr A and Mr B are discussing the merits of their cars. Mr A can go from 0 to 50 mph in ten seconds, and Mr B can go from 0 to 60 mph in 20 seconds. Mr B gives Mr A a start of one second. Assuming that each driver first accelerates uniformly to his maximum speed and thereafter travels at each uniform speed, how long does it take Mr B to catch Mr A, and how far have the cars travelled by then?

Answer: 41 s, half a mile.

I make the answers as follows. Let me know ( [email protected] ) if you think I have got any of them wrong.

Uniform Acceleration

• September 13, 2023
• Kinematics , Mechanics

Uniform acceleration is a specific type of motion in which an object’s velocity changes at a constant rate over time. It serves as a foundational concept in physics, particularly in the study of mechanics. The purpose of this article is to give readers a thorough understanding of uniform acceleration by looking into its definition, characteristics, kinematic equations, graphical representations, and real-world examples.

Table of Contents

Definition of Uniform Acceleration

If the velocity of a body changes by an equal amount in an equal interval of time, however, small the interval may be then its acceleration is said to be uniform.

Uniform acceleration (\( a \)) can also be defined as the rate of change of velocity per unit time, where this rate of change remains constant. In mathematical terms, uniform acceleration is defined by the equation:

\[ a = \frac{{\Delta v}}{{\Delta t}} \]

Here, \( \Delta v \) represents the change in velocity, and \( \Delta t \) denotes the change in time.

Characteristics of Uniform Acceleration

Understanding the nuances of uniform acceleration becomes easier when we delve into its core characteristics. Here is a detailed breakdown:

Constant Rate of Change in Velocity

In uniform acceleration, the velocity of an object changes at a consistent rate.

Applicability of Kinematic Equations

Uniformly accelerated motion can be described using three fundamental kinematic equations: 1. \( v = u + at \) 2. \( s = ut + \frac{1}{2}at^2 \) 3. \( v^2 = u^2 + 2as \)

Linear Velocity-Time Graph

A velocity-time graph for uniform acceleration is a straight line, and its slope equals the value of the constant acceleration.

Parabolic Displacement-Time Graph

The displacement-time graph for such motion is a parabola, arising from the quadratic relationship $(\frac{1}{2}at^2)$ between displacement and time in the second equation of motion.

The figure given below shows the distance-time graph of an object moving with a uniform accelerated motion where acceleration $a=2m/s^2$.

Importance of Initial Conditions

Initial velocity (\( u \)) and initial position (\( s_0 \)) are essential parameters for problem-solving in uniformly accelerated motion. These conditions act as the starting point for any calculations involving kinematic equations.

Idealized Conditions

It is often an idealized model but serves as a practical approximation for real-world applications like free-falling objects in a vacuum or a car accelerating on a straight road.

Directional Aspects

Both the velocity and acceleration vectors can be in the same or opposite directions, determining whether the object speeds up or slows down.

Kinematic Equations for Uniformly Accelerated Motion

Uniformly accelerated motion can be modeled using the following kinematic equations:

1. First Equation of Motion \[ v = u + at \]

2. Second Equation of Motion: \[ s = ut + \frac{1}{2} a t^2 \]

3. Third Equation of Motion: \[ v^2 = u^2 + 2as \]

These equations allow us to relate position, velocity, and acceleration to time, making them invaluable tools for problem-solving.

Graphical Interpretation

In uniformly accelerated motion, different graphs can portray the relationship between parameters:

• Velocity-Time Graph: A straight line indicates uniform acceleration, and its slope is equal to the value of the constant acceleration.
• Displacement-Time Graph: This graph is a parabola, highlighting the quadratic nature of the relationship between displacement and time.

Examples of Uniform Acceleration

Understanding uniform acceleration conceptually often benefits from specific examples. Below are points illustrating scenarios where uniform acceleration is observed or can be a reasonable approximation.

Falling Objects in a Vacuum

• Context: An object falling freely near the Earth’s surface in the absence of air resistance.
• Acceleration: The gravitational acceleration \( g = 9.81 \, \text{m/s}^2 \) is constant.
• Example Calculation: If an object starts from rest, its velocity after 2 seconds would be \( v = gt = 9.81 \times 2 = 19.62 \, \text{m/s} \).

Cars Accelerating from Rest

• Context: A car accelerates from rest to reach a certain speed on a straight road.
• Acceleration: Constant force from the engine results in constant acceleration.
• Example Calculation: With \( a = 2 \, \text{m/s}^2 \) and starting from rest (\( u = 0 \)), the car’s velocity after 3 seconds would be \( v = at = 2 \times 3 = 6 \, \text{m/s} \).

Inclined Plane without Friction

• Context: An object sliding down an inclined plane without friction.
• Acceleration: The component of the gravitational force along the incline provides a constant acceleration.
• Example Calculation: If the incline is at \( 30^\circ \) and neglecting friction, the acceleration \( a = g \sin(30^\circ) = 4.905 \, \text{m/s}^2 \).

Spacecraft Thrust

• Context: A spacecraft in outer space applying constant thrust.
• Acceleration: Due to the constant thrust and lack of air resistance in space, the acceleration remains constant.
• Example Calculation: With \( a = 0.5 \, \text{m/s}^2 \), the spacecraft’s velocity after 10 seconds from rest would be \( v = at = 0.5 \times 10 = 5 \, \text{m/s} \).

Elevator Moving Upward

Context: An elevator starts from rest and moves upward with constant acceleration. Acceleration: The motor provides a constant force, and thus, the acceleration is constant. Example Calculation: If \( a = 1 \, \text{m/s}^2 \), the elevator’s velocity after 4 seconds from rest would be \( v = at = 1 \times 4 = 4 \, \text{m/s} \).

Each of these examples can be analyzed using the kinematic equations for uniformly accelerated motion. They serve as practical applications of the concept, aiding in your understanding and problem-solving skills.

Questions and Answers

Is gravitational acceleration a form of uniform acceleration.

Yes, in a vacuum, gravitational acceleration is approximately constant near the Earth’s surface.

How can time \( t \) be calculated using the first equation of motion

Time can be determined by rearranging the first equation: \( t = \frac{v-u}{a} \).

Is uniform acceleration applicable in circular motion?

No, because the direction of acceleration changes continuously in circular motion.

By understanding its definitions, characteristics, and associated equations, you lay the groundwork for solving a wide range of problems in physics. Whether you’re observing a car accelerating on a straight road or an object free-falling in a vacuum, the concept of uniform acceleration frequently applies, making it essential for any student or practitioner of physics.

Further Reading

1. Kinematic Equations 2. Uniform Motion  3.  Concept of 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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UP Class 11 Physics

Course: up class 11 physics   >   unit 3.

• Choosing kinematic equations
• Choosing the best kinematic equation
• Using equations of motion (1 step numerical)

Using equations of motion (2 steps numerical)

• Kinematic equations: calculations
• Free fall - 2 body solved numerical
• Solving freefall problems using kinematic formulas

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Video transcript

3.4 Motion with Constant Acceleration

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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Access for free at https://openstax.org/books/university-physics-volume-1/pages/1-introduction

• 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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