2.8

Concepts & Calculations

In this chapter we have studied the displacement, velocity, and acceleration vectors. We conclude now by presenting several examples that review some of the important features of these concepts. The three-part format of these examples stresses the role of conceptual understanding in problem solving. First, the problem statement is given. Then, there is a concept question-and-answer section, followed by the solution section. The purpose of the concept question-and-answer section is to provide help in understanding the solution and to illustrate how a review of the concepts can help in anticipating some of the characteristics of the numerical answers.

| Concepts & Calculations Example 17 | | | Skydiving

A skydiver is falling straight down, along the negative y direction. (a) During the initial part of the fall, her speed increases from 16 to 28 m/s in 1.5 s, as in Figure 2-23a. (b) Later, her parachute opens, and her speed decreases from 48 to 26 m/s in 11 s, as in part b of the drawing. In both instances, determine the magnitude and direction of her average acceleration.

Figure 2-23 (a) A skydiver falls initially with her parachute unopened. (b) Later on, she opens her parachute. Her acceleration is different in the two parts of the motion. The initial and final velocities are v0 and v, respectively.

Concept Questions and Answers Is her average acceleration positive or negative when her speed is increasing?

Answer Since her speed is increasing, the acceleration vector must point in the same direction as the velocity vector, which points in the negative y direction. Thus, the acceleration is negative.

Is her average acceleration positive or negative when her speed is decreasing?

Answer Since her speed is decreasing, the acceleration vector must point opposite to the velocity vector. Since the velocity vector points in the negative y direction, the acceleration must point in the positive y direction. Thus, the acceleration is positive.

Solution

a. | Since the skydiver is moving in the negative y direction, her initial velocity is v0 = −16 m/s and her final velocity is v = −28 m/s. Her average acceleration is the change in the velocity divided by the elapsed time:

| (2.4)

As expected, her average acceleration is negative. Note that her acceleration is not that due to gravity (−9.8 m/s2) because of wind resistance.
b. | Now the skydiver is slowing down, but still falling along the negative y direction. Her initial and final velocities are v0 = − 48 m/s and v = −26 m/s, respectively. The average acceleration for this phase of the motion is

| (2.4)

Now, as anticipated, her average acceleration is positive.

| Concepts & Calculations Example 18 | | | A Top-Fuel Dragster

A top-fuel dragster starts from rest and has a constant acceleration of 40.0 m/s2. What are the (a) final velocities and (b) displacements of the dragster at the end of 2.0 s and at the end of twice this time, or 4.0 s? Concept Questions and Answers At a time t the dragster has a certain velocity. When the time doubles to 2t, does the velocity also double?

Answer Because the dragster has an acceleration of 40.0 m/s2, its velocity changes by 40.0 m/s during each second of travel. Therefore, since the dragster starts from rest, the velocity is 40.0 m/s at the end of the 1st second, 2 × 40.0 m/s at the end of the 2nd second, 3 × 40.0 m/s at the end of the 3rd second, and so on. Thus, when the time doubles, the velocity also doubles.

When the time doubles to 2t, does the displacement of the dragster also double?

Answer The displacement of the dragster is equal to its average velocity multiplied by the elapsed time. The average velocity is just one-half the sum of the initial and final velocities, or . Since the initial velocity is zero, v0 = 0 m/s and the average velocity is just one-half the final velocity, or . However, as we have seen, the final velocity is proportional to the elapsed time, since when the time doubles, the final velocity also doubles. Therefore, the displacement, being the product of the average velocity and the time, is proportional to the time squared, or t2. Consequently, as the time doubles, the displacement does not double, but increases by a factor of four.

Solution

a. | According to Equation 2.4, the final velocity v, the initial velocity v0, the acceleration a, and the elapsed time t are related by . The final velocities at the two times are

We see that the velocity doubles when the time doubles, as expected.
b. | The displacement x is equal to the average velocity multiplied by the time, so

where we have used the fact that v0 = 0 m/s. According to Equation 2.4, the final velocity is related to the acceleration by , or v = at, since v0 = 0 m/s. Therefore, the displacement can be written as . The displacements at the two times are then

As predicted, the displacement at t = 4.0 s is four times the displacement at t = 2.0 s.

At the end of the problem set for this chapter, you will find homework problems that contain both conceptual and quantitative parts. These problems are grouped under the heading Concepts & Calculations.

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