2.5

Applications of the Equations of Kinematics

The equations of kinematics can be applied to any moving object, as long as the acceleration of the object is constant. However, remember that each equation contains four variables. Therefore, numerical values for three of the four must be available if an equation is to be used to calculate the value of the remaining variable. To avoid errors when using these equations, it helps to follow a few sensible guidelines and to be alert for a few situations that can arise during your calculations.
Decide at the start which directions are to be called positive (+) and negative (−) relative to a conveniently chosen coordinate origin. This decision is arbitrary, but important because displacement, velocity, and acceleration are vectors, and their directions must always be taken into account. In the examples that follow, the positive and negative directions will be shown in the drawings that accompany the problems. It does not matter which direction is chosen to be positive. However, once the choice is made, it should not be changed during the course of the calculation.
As you reason through a problem before attempting to solve it, be sure to interpret the terms “decelerating” or “deceleration” correctly, should they occur in the problem statement. These terms are the source of frequent confusion, and Conceptual Example 7 offers help in understanding them.

| Conceptual Example 7 | | | Deceleration Versus Negative Acceleration

A car is traveling along a straight road and is decelerating. Does the car’s acceleration a necessarily have a negative value? Reasoning and Solution We begin with the meaning of the term “decelerating,” which has nothing to do with whether the acceleration a is positive or negative. The term means only that the acceleration vector points opposite to the velocity vector and indicates that the moving object is slowing down. When a moving object slows down, its instantaneous speed (the magnitude of the instantaneous velocity) decreases. One possibility is that the velocity vector of the car points to the right, in the positive direction, as Figure 2-11a shows. The term “decelerating” implies that the acceleration vector points opposite, or to the left, which is the negative direction. Here, the value of the acceleration a would indeed be negative. However, there is another possibility. The car could be traveling to the left, as in Figure 2-11b. Now, since the velocity vector points to the left, the acceleration vector would point opposite or to the right, according to the meaning of the term “decelerating.” But right is the positive direction, so the acceleration a would have a positive value in Figure 2-11b. We see, then, that a decelerating object does not necessarily have a negative acceleration.

Figure 2-11 When a car decelerates along a straight road, the acceleration vector points opposite to the velocity vector, as Conceptual Example 7 discusses.

Related Homework: Problems 38, 64

Sometimes there are two possible answers to a kinematics problem, each answer corresponding to a different situation. Example 8 discusses one such case.

Need more practice?

Interactive LearningWare2.1
Here we revisit Example 5, in which the positive direction was assumed to point toward the right (see Figure 2-9). Now, however, we assume that the positive direction points to the left and will show how to deal with such a choice and how to interpret the results.

| Example 8 | | | An Accelerating Spacecraft

The spacecraft shown in Figure 2-12a is traveling with a velocity of +3250 m/s. Suddenly the retrorockets are fired, and the spacecraft begins to slow down with an acceleration whose magnitude is 10.0 m/s2. What is the velocity of the spacecraft when the displacement of the craft is +215 km, relative to the point where the retrorockets began firing?

Figure 2-12 (a) Because of an acceleration of −10.0 m/s2, the spacecraft changes its velocity from v0 to v. (b) Continued firing of the retrorockets changes the direction of the craft’s motion.

The physics of the acceleration caused by a retrorocket.

Reasoning Since the spacecraft is slowing down, the acceleration must be opposite to the velocity. The velocity points to the right in the drawing, so the acceleration points to the left, in the negative direction; thus, a = −10.0 m/s2. The three known variables are listed as follows:

Spacecraft Data
x | a | v | v0 | t
+215 000 m | −10.0 m/s2 | ? | +3250 m/s |

The final velocity v of the spacecraft can be calculated using Equation 2.5, since it contains the four pertinent variables.

Need more practice?

Interactive LearningWare2.2
A bus has stopped to pick up riders. Awoman is running at a constant velocity of +5.0 m/s in an attempt to catch the bus. When she is 11 m from the bus, it pulls away with a constant acceleration of +1.0 m/s2. From this point, how much time does it take her to reach the bus if she keeps running with the same velocity?
Related Homework: Problems 28, 77, 78

Solution From Equation 2.9 (), we find that

Both of these answers correspond to the same displacement (x = +215 km), but each arises in a different part of the motion. The answer v = +2500 m/s corresponds to the situation in Figure 2-12a, where the spacecraft has slowed to a speed of 2500 m/s, but is still traveling to the right. The answer v = −2500 m/s arises because the retrorockets eventually bring the spacecraft to a momentary halt and cause it to reverse its direction. Then it moves to the left, and its speed increases due to the continually firing rockets. After a time, the velocity of the craft becomes v = −2500 m/s, giving rise to the situation in Figure 2-12b. In both parts of the drawing the spacecraft has the same displacement, but a greater travel time is required in part b compared to part a.
The motion of two objects may be interrelated, so they share a common variable. The fact that the motions are interrelated is an important piece of information. In such cases, data for only two variables need be specified for each object. See Interactive Learning Ware 2.2 for an example that illustrates this point.
Often the motion of an object is divided into segments, each with a different acceleration. When solving such problems, it is important to realize that the final velocity for one segment is the initial velocity for the next segment, as Example 9 illustrates.

A N A L Y Z I N G M U L T I P L E - C O N C E P T P R O B L E M S

| Example 9 | | | A Motorcycle Ride

A motorcycle ride consists of two segments, as shown in Figure 2-13a. During the first segment, the motorcycle starts from rest, has an acceleration of +2.6 m/s2 and a displacement of +120 m. Immediately after the first segment, the motorcycle enters the second segment and begins slowing down with an acceleration of −1.5 m/s2 until its velocity is +12 m/s. What is the displacement of the motorcycle during the second segment?

Figure 2-13 (a) This motorcycle ride consists of two segments, each with a different acceleration. (b) The variables for the second segment. (c) The variables for the first segment.

Reasoning We can use an equation of kinematics from Table 2-1 to find the displacement x2nd for the second segment. To do this, it will be necessary to have values for three of the variables that appear in the equation. Values for the acceleration () and final velocity () are given. A value for a third variable, the initial velocity v0, 2nd, can be obtained by noting that it is also the final velocity of the first segment. The final velocity of the first segment can be found by using an appropriate equation of kinematics, since three variables (x1st, a1st, and v0, 1st) are known for this part of the motion, as the following table reveals.
Knowns and Unknowns The data for this problem are listed in the table:

Description | Symbol | Value | Comment
Explicit Data
Displacement for 1st segment | x1st | +120 m |
Acceleration for 1st segment | a1st | +2.6 m/s2 | Positive, because the motorcycle moves in the +x direction and speeds up.
Acceleration for 2nd segment | a2nd | −1.5 m/s2 | Negative, because the motorcycle moves in the +x direction and slows down.
Final velocity for 2nd segment | v2nd | +12 m/s |
Implicit Data
Initial velocity for 1st segment | v0, 1st | 0 m/s | The motorcycle starts from rest.
Unknown Variable
Displacement for 2nd segment | x2nd | ? |

Modeling the Problem

| Step 1 2nd Segment Displacement
| Figure 2-13b shows the situation during the 2nd segment. Two of the variables—the final velocity v2nd and the acceleration a2nd—are known, and for convenience we choose Equation 2.9 to find the displacement x2nd of the motorcycle:

Solving this relation for x2nd yields Equation 1 at the right. Although the initial velocity V0,2nd of the 2nd segment is not known, we will be able to determine its value from a knowledge of the motion during the 1st segment, as outlined in Steps 2 and 3.

| (1)

| Step 2 Initial Velocity of the 2nd Segment
| Since the motorcycle enters the 2nd segment immediately after leaving the 1st segment, the initial velocity v0,2nd of the 2nd segment is equal to the final velocity v1st of the 1st segment, or v0,2nd = v1st. Squaring both sides of this equation gives

This result can be substituted into Equation 1 as shown at the right. In the next step v1st will be determined.

| Step 3 Final Velocity of 1st Segment
| Figure 2-13c shows the motorcycle during the 1st segment. Since we know the initial velocity v0, 1st, the acceleration a1st, and the displacement x1st, we can employ Equation 2.9 to find the final velocity v1st at the end of the 1st segment:

This relation for v1st2 can be substituted into Equation 2, as shown at the right.

Solution Algebraically combining the results of each step, we find that

The displacement x2nd of the motorcycle during the 2nd segment is

Related Homework: Problems 32, 52, 72

Now that we have seen how the equations of kinematics are applied to various situations, it’s a good idea to summarize the reasoning strategy that has been used. This strategy, which is outlined below, will also be used when we consider freely falling bodies in Section 2.6 and two-dimensional motion in Chapter 3.

| Reasoning Strategy
| Applying the Equations of Kinematics

1. | Make a drawing to represent the situation being studied. A drawing helps us to see what’s happening.
2. | Decide which directions are to be called positive (+) and negative (−) relative to a conveniently chosen coordinate origin. Do not change your decision during the course of a calculation.
3. | In an organized way, write down the values (with appropriate plus and minus signs) that are given for any of the five kinematic variables (x, a, v, v0, and t). Be on the alert for “implicit data,” such as the phrase “starts from rest,” which means that the value of the initial velocity is v0 = 0 m/s. The data summary boxes used in the examples in the text are a good way of keeping track of this information. In addition, identify the variables that you are being asked to determine.
4. | Before attempting to solve a problem, verify that the given information contains values for at least three of the five kinematic variables. Once the three known variables are identified along with the desired unknown variable, the appropriate relation from Table 2.1 can be selected. Remember that the motion of two objects may be interrelated, so they may share a common variable. The fact that the motions are interrelated is an important piece of information. In such cases, data for only two variables need be specified for each object.
5. | When the motion of an object is divided into segments, as in Example 9, remember that the final velocity of one segment is the initial velocity for the next segment.
6. | Keep in mind that there may be two possible answers to a kinematics problem as, for instance, in Example 8. Try to visualize the different physical situations to which the answers correspond.

As this bald eagle comes in for a landing, its velocity vector points in the same direction as its motion. However, since the eagle is slowing down, its acceleration vector points opposite to the velocity vector. (Tom & Pat Lesson/Photo Researchers, Inc.)

Test your understanding of the equations of kinematics for constant acceleration, which are discussed in Sections 2.4 and 2.5.

Copyright © 2007 John Wiley & Sons, Inc. All rights reserved.