# Homework, Chapter 3: Linear Motion

Ch 3, Linear Motion; Ex 22, 24, 27, 34, 37; Pb 2, 3, 4, 6

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## Exercises (Discussion Questions)

Ex 3.22 Suppose that a freely falling object were somehow equipped with a speedometer. By how much would its speed reading increase with each second of fall?

Every second, its speed increases by 10 m/s. That means its acceleration is 10 (m/s)/s.

Ex 3.24 For a freely falling object dropped from rest, what is its acceleration at the end of the 5th second of fall? The 10th second? Defend your answer.

The acceleration of a freely falling object is constant so its acceleration at the end of the 5th second is the same as its acceleration at the end of the 10th second--or the end of the first second or the beginning of the 17th second. Its acceleration is 9.8 m/s/s -- which we will approximate as nearly 10 m/s/s.

Ex 3.27 Someone standing at the edge of a cliff (as in Figure 3.8) throws a ball straight up at a certain speed and another ball straight down with the same initial speed. If air resistance is negligible, which ball will have the greater speed when it strikes the ground below? Explain.

First, consider the ball thrown up. It continues to move up as it slows and then comes to a stop at the top of its path and then starts to increase its speed downward. When it gets back to its original position at the edge of the cliff, it has the same speed as it did initially when it was thrown up; it is moving in the opposite direction (downward, this time) but has the same speed. It will continue on and hit the ground below. Its motion from the cliff on down to the ground, then, will be exactly the same as the motion of a ball thrown down with this same speed.

That means the two balls hit the ground with the same speed.

Ex 3.34 Extend Tables 3.2 and 3.3 (which give values of time from 0 to 5 s) to 0 to 10 s, assuming no air resistance.

 time velocity distance fallen ( seconds) ( m/s ) ( meters ) 0 0 0 1 10 5 2 20 20 3 30 45 4 40 80 5 50 125 6 60 180 7 70 245 8 80 320 9 90 405 10 100 500 t (10) t (1/2) (10) t2

Ex 3.37 In this chapter, we studied idealized cases of balls rolling down smooth planes and objects falling with no air resistance. If a classmate complains that all this attention focused on idealized cases is valueless because idealized cases simply don't occur in the everyday world, how would you respond? How do you suppose the author of this book or the instructor of this course would respond?

The motion of idealized systems is far easier to understand. Once that motion is understood--as a first approximation--the motion with friction (such as air resistance) is then easier to understand as a refinement to the first approximation.

## (Numerical) Problems

Pb 3.2 What is the acceleration of a vehicle that changes its velocity from 100 km/h to a dead stop in 10 s

The change in velocity is v = vf - vi = 0 - 100 km/h = - 100 km/h. The minus sign is important; don't forget to include it. The acceleration is a = v / t = ( - 100 km / h ) / 10 s = - 10 (km/h) / s or a = - 10 km/h/s.

Pb 3.3 A ball is thrown straight up with an initial speed of 30 m/s. How high does it go, and how long is it in the air (neglecting air resistance)?

It is easier to begin by asking "how long is it in the air?" On the way up, its speed decreases by 10 m/s. At the end of the first second it is moving up at 20 m/s. At the end of the second second, it is moving up at 10 m/s. At the end of the third second, its speed (and velocity) is (or are) zero; it stops for just an instant. So it takes three seconds to go up to its highest position. The motion is symmetric. It will take and additional three seconds to fall back down to its original position.

So it is in the air for a total of six seconds.

Now we can ask "how high does it go?" It may be easier to think of this as "how far does it fall in three seconds?" It falls from rest and we have developed the equation s = 1/2 a t2 or y = 1/2 a t2 so we can use that

y = 1/2 a t2

y = 1/2 (10 m/s2) ( 3 s )2

y = 45 m

Pb 3.4 A ball is thrown with enough speed straight up so that it is in the air several seconds.

(a) What is the velocity of the ball when it gets to its highest point?

At the top, its velocity is zero.

(b) What is its velocity 1 s before it reaches its highest point?

One second before reaching the top, it is moving up at v = 10 m/s.

(c) What is the change in its velocity during this 1-s interval?

v = vf - vi = 0 - 10 m/s = - 10 m/s

(d) What is its velocity 1 s after it reaches its highest point?

One second after reaching the top, it is moving down a v = - 10 m/s; the minus sign on the velocity indicates that it is, indeed, moving down.

(e) What is the change in velocity during this 1-s interval?

v = vf - vi = 10 m/s - 0 = - 10 m/s

(f) What is the change in velocity during the 2-s interval?

v = vf - vi = - 10 m/s - 10 m/s = - 20 m/s

(g) What is the acceleration of the ball during any of these time intervals and when it passes through the zero velocity point?

a = v / t

For part (c), this is a = v / t = ( - 10 m/s) / (1 s) = - 10 m/s2.

For part (e), this is a = v / t = ( - 10 m/s) / (1 s) = - 10 m/s2.

For part (d), this is a = v / t = ( - 20 m/s) / (2 s) = - 10 m/s2.

This illustrates that the acceleration is constant; the acceleration is always a = - 10 m/s2. That is true on the way up, on the way down, and even at the moment the ball is at the very top of its path.

Pb 3.6 A car takes 10 s to go from v = 0 to v = 30 m/s at approximately constant acceleration. If you wish to find the distance traveled using the quation d = (1/2) a t2, what value should you use for a?

Acceleration a is the change in velocity divided by the change in time,

a = v / t

v = vfinal - vinitial

v =30 m/s - 0 = 30 m/s

t = 10 s

a = v / t = [30 m/s] / 10 s

a = 3 m/s/s

a = 3 m/s2

Let's go ahead and actually calculate how far the car travels during that 10 time;

d = (1/2) a t2

d = (0.5)(3 m/s2)(10 s)2

d = (0.5)(3 m/s2)(100 s2)

d = 150 m

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Typical or possible multiple-choice questions over this material:

1. Kinematics is a description of motion. Motion was first well understood

a) by Aristotle and the ancient Greeks

b) by Ptolemy in Egypt

c) by Galileo in Italy

d) not until the beginning of the twentieth century

2. To measure the time needed to investigate motion,

a) Aristotle used the pendulum clock which had just been invented

b) Ptolemy used a sundial

c) Galileo invented his own water clocks

d) Newton invented the pendulum clock

3. What is the average speed of a motorcycle that travels 20 m in 2 s?

a) 40 m/s

b) 20 m/s

c) 10 m/s

d) 9.8 m/s

4. What is the average speed of a car that travels 45 km in 3 h?

a) 135 km/h

b) 15 km/h

c) 10 km/h

d) 9.8 km/h

5. Consider a train that has an acceleration of 3 m/s2. Initially, at time t = 0, it has a velocity of vi = 10 m/s. What is its speed at t = 3 s?

a) 40 m/s

b) 30 m/s

c) 23 m/s

d) 19 m/s

6. Consider a car that starts at rest and accelerates at 2 m/s2 for 3 seconds. At that time, t = 3 s, how fast is it going?

a) 12 m/s

b) 9 m/s

c) 6 m/s

d) 3 m/s

7. Consider a car that starts at rest and accelerates at 2 m/s2 for 3 seconds. At that time, t = 3 s, how far has it gone?

a) 12 m

b) 9 m

c) 6 m

d) 3 m

8. Consider a ball that is thrown upward at the edge of a canyon with an initial velocity of 20 m/s. Three seconds later, what is its velocity?

a) 30 m/s

b) 15 m/s

c) - 10 m/s

d) - 30 m/s

9. Consider a ball that is thrown straight upward at the edge of a canyon with an initial velocity of 20 m/s. Three seconds later, where is it located? Take its initial position, at the edge of the canyon, to be the origin; that is, yi = 0.

a) 30 m

b) 15 m

c) - 10 m

d) - 30 m

1. Kinematics is a description of motion. Motion was first well understood

a) by Aristotle and the ancient Greeks

b) by Ptolemy in Egypt

c) by Galileo in Italy

d) not until the beginning of the twentieth century

2. To measure the time needed to investigate motion,

a) Aristotle used the pendulum clock which had just been invented

b) Ptolemy used a sundial

c) Galileo invented his own water clocks

d) Newton invented the pendulum clock

3. What is the average speed of a motorcycle that travels 20 m in 2 s?

a) 40 m/s

b) 20 m/s

c) 10 m/s; v = 20 m / 2 s = 10 m/s

d) 9.8 m/s

4. What is the average speed of a car that travels 45 km in 3 h?

a) 135 km/h

b) 15 km/h; v = 45 km / 3 h = 15 km/h

c) 10 km/h

d) 9.8 km/h

5. Consider a train that has an acceleration of 3 m/s2. Initially, at time t = 0, it has a velocity of vi = 10 m/s. What is its speed at t = 3 s?

a) 40 m/s

b) 30 m/s

c) 23 m/s

d) 19 m/s; v = vi + a t = 10 m/s + (3 m/s2) (3 s) = (10 + 9) m/s = 19 m/s

6. Consider a car that starts at rest and accelerates at 2 m/s2 for 3 seconds. At that time, t = 3 s, how fast is it going?

a) 12 m/s

b) 9 m/s

c) 6 m/s; v = vi + a t = 0 + (2 m/s2) (3 s) = 6 m/s

d) 3 m/s

7. Consider a car that starts at rest and accelerates at 2 m/s2 for 3 seconds. At that time, t = 3 s, how far has it gone?

a) 12 m

b) 9 m; x = xi + vi t + (1/2) a t2 = 0 + 0 + (1/2) (2 m/s2) (3 s)2 = 9 m

c) 6 m

d) 3 m

8. Consider a ball that is thrown upward at the edge of a canyon with an initial velocity of 20 m/s. Three seconds later, what is its velocity?

a) 30 m/s

b) 15 m/s

c) - 10 m/s; v = vi + a t = 20 m/s + ( - 10 m/s2) (3 s) = (20 - 30) m/s = - 10 m/s

d) -- 30 m/s

9. Consider a ball that is thrown straight upward at the edge of a canyon with an initial velocity of 20 m/s. Three seconds later, where is it located? Take its initial position, at the edge of the canyon, to be the origin; that is, yi = 0.

a) 30 m

b) 15 m; y = yi + vi t + (1/2) a t2 = 0 + (20 m/s) (3 s) + (1/2) (- 10 m/s2) (3 s)2 = (0 + 60 - 45) m = 15 m

c) - 10 m

d) - 30 m

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