Chapter 2 Solutions
Kinematic Vocabulary
One of the difficulties in studying mechanics is that many common words are used with highly
specific technical meanings, among them velocity, acceleratio n, position, speed, and
displacement. The series of questions in this problem is designed to get you to try to think of
these quantities like a physicist.
Answer the questions in this problem using words from the following list:
A.
B.
C.
D.
E.
F.
G.
H.
I.
J.
K.
position
direction
displacement
coordinates
velocity
acceleration
distance
magnitude
vector
scalar
components
Part A
Velocity differs from speed in that velocity indicates a particle's __________ of motion.
Enter the letter from the list given in the problem introduction that best completes the sentence.
ANSWER:
B
Also accepted: direction
Part B
Unlike speed, velocity is a __________ quantity.
Enter the letter from the list given in the problem introduction that best completes the sentence.
ANSWER:
I
Also accepted: vector
Part C
A vector has, by definition, both __________ and direction.
Enter the letter from the list given in the problem introduction that best completes the sentence.
ANSWER:
H
Also accepted: magnitude
Part D
Once you have selected a coordinate system, you can express a two-dimensional vector using a
pair of quantities known collectively as __________.
Enter the letter from the list given in the problem introduction that best completes the sentence.
ANSWER:
K
Also accepted: components, D, coordinates
Part E
Speed differs from velocity in the same way that __________ differs from displacement.
Enter the letter from the list given in the problem introduction that best completes the sentence.
Hint 1. Definition of displacement
Displacement is the vector that indicates the difference of two positions (e.g., the final position
from the initial position). Being a vector, it is independent of the coordinate system used to
describe it (although its vector components depend on the coordinate system).
ANSWER:
G
Also accepted: distance
Part F
Consider a physical situation in which a particle moves from point A to point B. This process is
described from two coordinate systems that are identical except that they have different origins.
The __________ of the particle at point A differ(s) as expressed in one coordinate system
compared to the other, but the __________ from A to B is/are the same as expressed in both
coordinate systems.
Type the letters from the list given in the problem introduction that best complete the sentence.
Separate the letters with commas. There is more than one correct answer, but you should only
enter one pair of comma-separated letters. For example, if the words "vector" and "scalar" fit
best in the blanks, enter I,J.
ANSWER:
A,C
Also accepted: position,C, D,C, coordinates,C, A,displacement, position,displacement,
D,displacement, coordinates,displacement, A,G, position,G, D,G, coordinates,G, A,distance,
position,distance, D,distance, coordinates,distance, A,E, position,E, D,E, coordinates,E,
A,velocity, position,velocity, D,velocity, coordinates,velocity, A,B, position,B, D,B,
coordinates,B, A,direction, position,direction, D,direction, coordinates,direction
The coordinates of a point will depend on the coordinate system that is chosen, but there are
several other quantities that are independent of the choice of origin for a coordinate system: in
particular, distance, displacement, direction, and velocity. In working physics problems, unless
you are interested in the position of an object or event relative to a specific origin, you can
usually choose the coordinate system origin to be wherever is most convenient or intuitive.
Note that the vector indicating a displacement from A to B is usually represented as
.
Part G
Identify the following physical quantities as scalars or vectors.
ANSWER:
Direction of Velocity and Acceleration Vector Quantities Conceptual Question
For each of the motions described below, determine the algebraic sign ( , , or ) of the
velocity and acceleration of the object at the time specified. For all of the motions, the positive y
axis is upward.
Part A
An elevator is moving downward when someone presses the emergency stop button. The
elevator comes to rest a short time later. Give the signs for the velocity and the acceleration of
the elevator after the button has been pressed but before the elevator has stopped.
Enter the correct sign for the elevator's velocity and the correct sign for the elevator's
acceleration, separated by a comma. For example, if you think that the velocity is positive and
the acceleration is negative, then you would enter +,- . If you think that both are zero, then you
would enter 0,0 .
Hint 1. Algebraic sign of velocity
The algebraic sign of velocity is determined solely by comparing the direction in which the
object is moving with the direction that is defined to be positive. In this example, upward is
defined to be positive. Therefore, any object moving upward, whether speeding up, slowing
down, or traveling at constant speed, has positive velocity.
Hint 2. Algebraic sign of acceleration
The algebraic sign of acceleration is more difficult to determine than the algebraic sign of
velocity. The acceleration of an object points in the same direction as the change in the velocity
of an object. If an object is speeding up, the change in the velocity points in the same direction as
the velocity:
If an object is slowing down, the change in velocity points in the opposite direction to that of the
velocity:
Once you know the direction of the acceleration, you can determine its sign by comparing it to
the defined positive direction, in this case, upward.
ANSWER:
-,+
Part B
A child throws a baseball directly upward. What are the signs of the velocity and acceleration of
the ball immediately after the ball leaves the child's hand?
Enter the correct sign for the baseball's velocity and the correct sign for the baseball's
acceleration, separated by a comma. For example, if you think that the velocity is positive and
the acceleration is negative, then you would enter +,- . If you think that both are zero, then you
would enter 0,0 .
Hint 1. Algebraic sign of velocity
The algebraic sign of velocity is determined solely by comparing the direction in which the
object is moving with the direction that is defined to be positive. In this example, upward is
defined to be positive. Therefore, any object moving upward, whether speeding up, slowing
down, or traveling at constant speed, has positive velocity.
Hint 2. Algebraic sign of acceleration
The algebraic sign of acceleration is more difficult to determine than the algebraic sign of
velocity. The acceleration of an object points in the same direction as the change in the velocity
of an object. If an object is speeding up, the change in the velocity points in the same direction as
the velocity:
If an object is slowing down, the change in velocity points in the opposite direction to that of the
velocity:
Once you know the direction of the acceleration, you can determine its sign by comparing it to
the defined positive direction, in this case, upward.
ANSWER:
+,-
Part C
A child throws a baseball directly upward. What are the signs of the velocity and acceleration of
the ball at the very top of the ball's motion (i.e., the point of maximum height)?
Enter the correct sign for the baseball's velocity and the correct sign for the baseball's
acceleration, separated by a comma. For example, if you think that the velocity is positive and
the acceleration is negative, then you would enter +,- . If you think that both are zero, then you
would enter 0,0 .
Hint 1. Algebraic sign of velocity
The algebraic sign of velocity is determined solely by comparing the direction in which the
object is moving with the direction that is defined to be positive. In this example, upward is
defined to be positive. Therefore, any object moving upward, whether speeding up, slowing
down, or traveling at constant speed, has positive velocity.
Hint 2. Algebraic sign of acceleration
The algebraic sign of acceleration is more difficult to determine than the algebraic sign of
velocity. The acceleration of an object points in the same direction as the change in the velocity
of an object. If an object is speeding up, the change in the velocity points in the same direction as
the velocity:
If an object is slowing down, the change in velocity points in the opposite direction as the
velocity:
Once you know the direction of the acceleration, you can determine its sign by comparing it to
the defined positive direction, in this case, upward.
ANSWER:
0,-
Solutions to Problems
2.2. Set Up: From the graph the position xt at each time t is: x1  10 m, x2  0, x3   10 m, x4  0, x8  60 m, and
x10  60 m
Solve: (a) The displacement is x (i) x  x10  x1  50 m; (ii) x  x10  x3   70 m; (iii) x  x3  x2   10 m; (iv)
x  x4  x2  0
(b) (i) 30 m  10 m  40 m; 90 (ii) 10 m  10 m  20 m; (iii) zero (stays at x  60 m )
*2.11. Set Up: 10 century  100 yr 1 km  105 cm
Solve: (a) d  t  (50 cm/yr)(100 yr)  500 cm  50 m
(b) t 
d 550  105 cm

 11  107 yr
t
50 cm/yr
2.12. Set Up: The distance around the circular track is d   (400 m)  126 m For a half-lap, d  63 m Use coordinates
for which the origin is at her starting point and the x-axis is along a diameter, as shown in the figure below.
Solve: (a) After one lap she has returned to her starting point. Thus, x  0 and av, x  0
d 126 m

 201 m/s
t 625 s
x 400 m
d
63 m


 139 m/s; average speed  
 220 m/s
t
287 s
t 287 s
average speed 
(b) x  400 m and av, x
2.13. Set Up: Since sound travels at a constant speed, x  vxt; also, from the appendix we find that 1 mile is 1.609
km.
1 mi


Solve: x  (344 m/s)(75 s) 
  16 mi
 1609 103 m 
1 mi

 1
Reflect: The speed of sound is (344 m/s) 
  mi/s
3
 1609  10 m  5
185 m
d
185 m
 00243 s; pain: t 
 303 s
2.14. Solve: (a) t   touch: t 
762 m/s
0610 m/s

(b) The difference between the two times in (a) is 3.01 s.
2.31. Set Up: Assume the ball starts from rest and moves in the  x-direction We may use the equations for constant
acceleration.
Solve: (a) x  x0  150 m, vx  450 m/s and v0 x  0. Using vx 2  v0 x 2  2ax ( x  x0 ) gives
ax 
vx 2  v0 x 2 (450 m/s)2

 675 m/s2 .
2( x  x0 )
2(150 m)
2( x  x0 ) 2(150 m)
v v 

 00667 s
(b) Using x  x0   0 x x  t gives t 
v0 x  vx
450 m/s
2


Reflect: We could also use vx  v0 x  axt to find t 
vx 450 m/s

 00667 s, which agrees with our previous result.
ax 675 m/s2
The acceleration of the ball is very large.
2.34. Set Up: Take the  x direction to be the direction of motion of the boulder.
Solve: (a) Use the motion during the first second to find the acceleration. 0 x  0, x0  0, x  200 m, and t  100 s
x  x0  0 xt  12 axt 2 and ax 
2x
t
2

2(200 m)
(100 s)2
 400 m/s2
x  0 x  axt  (400 m/s2 )(100 s)  400 m/s
For the second second, 0 x  400 m/s, ax  400 m/s2 , and t  100 s
x  x0  0 xt  12 axt 2  (400 m/s)(100 s)  12 (400 m/s2 )(100 s)2  600 m
We can also solve for the location at t  200 s, starting at t  0 :
x  x0  0 xt  12 axt 2  12 (400 m/s2 )(200 s)2  800 m,
which agrees with 2.00 m in the first second and 6.00 m in the second second. The boulder speeds up so it travels farther
in each successive second.
(b) We have already found  x  400 m/s after the first second. After the second second,
x  0 x  axt  400 m/s  (400 m/s2 )(100 s)  800 m/s
*2.37. Set Up: Let  x be the direction the car is moving. We can use the equations for constant acceleration.
Solve: (a) From Eq. (2.13), with v0 x  0, ax 
vx 2
(20m/s)2

 167m/s2 
2( x  x0 ) 2(120m)
(b) Using Eq. (2.14), t  2( x  x0 )/vx  2(120m)/(20m/s)  12s
(c) (12 s)(20m/s)  240 m
Reflect: The average velocity of the car is half the constant speed of the traffic, so the traffic travels twice as far.
2.38. Set Up: 1 mi/h  1466 ft/s The car travels at constant speed during the reaction time. Let  x be the direction the
car is traveling, so ax   120 ft/s2 after the brakes are applied.
 1466 ft/s 
Solve: (a) 0 x  (150 mi/h) 
  220 ft/s During the reaction time the car travels a distance of (220 ft/s)
 1 mi/h 
(07 s)  154 ft.
For the motion after the brakes are applied, 0 x  220 ft/s, ax   120 ft/s2 , and  x  0  x 2  0 x 2  2ax ( x  x0 ) gives
( x  x0 ) 
 x 2  0 x 2
0  (220 ft/s)2
 202 ft
2(120 ft/s 2 )
The total distance is 154 ft  202 ft  356 ft
2a x

1466 ft/s 
  806 ft/s A calculation similar to that of part (a) gives a total stopping
 1 mi/h 
(b) 0 x  (550 mi/h) 
distance of ( x  x0 )  564 ft  2707 ft  327 ft
2.40. Set Up: Let  x be the direction the train is traveling. Find x  x0 for each segment of the motion.
Solve: t  0 to 14.0 s: x  x0  0 xt  12 axt 2  12 (160 m/s2 )(140 s)2  157 m At t  140 s, the speed is
x  0 x  ax  (160 m/s2 )(140 s)  224 m/s
In the next 70.0 s, ax  0 and x  x0  0 xt  (224 m/s)(700 s)  1568 m For the interval during which the train is
slowing down, 0 x  224 m/s, ax   350 m/s2 and  x  0  x 2  0 x 2  2ax ( x  x0 ) gives
x  x0 
 x 2  0 x 2

0  (224 m/s)2
2(350 m/s 2 )
The total distance traveled is 157 m  1568 m  72 m  1800 m
2a x
 72 m