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tiL12214_appe_643-698.indd Page 647 09/10/10 8:37 AM user-f463 Volume/207/MHDQ243/tiL12214_disk1of1/0073512214/tiL12214_pagefiles

d

All signifi cant figures are retained here because the units are − _ − m _

(c) v = ∴ d = v t = ( 20 ) (4 s)

s

defined exactly, without uncertainty. t
m _

2.15. The initial velocity (v i ) is given as 100.0 m/s, the fi nal velocity = 20 × 4 × s
s
(v f ) is given as 51.0 m/s, and the time is given as 5.00 s. _
m·s

Acceleration, including a deceleration or negative acceleration, = 80
s
is found from a change of velocity during a given time. Th us,
= 80 m
_
v f – v i
a = 2.18. Note that this problem can be solved with a series of three steps
t
as in the previous problem. It can also be solved by the equation
m _ m _

( 51.0 ) – ( 100.0 ) that combines all the relationships into one step. Either method
__
s
s

= is acceptable, but the following example of a one-step solution
5.00 s
reduces the possibilities of error since fewer calculations are
m _ involved:
– 49.0

_
s

=
1
5.00 s 1 _ 2 _ m _ 2
2 (
2)

d = gt = 9.8 (5.00 s)

m _ _ 2 s
1

= – 9.80 ×

s s 1 _ m _
2)
2 (
2

= 9.8 (25.0 s )
m _ s

= – 9.80
s 2 1 _ m _
( )

= (9.8)(25.0) × s 2
2
(The negative sign means a negative acceleration, or deceleration.) 2 s

_ 2
m·s
2.16. A ball thrown straight up decelerates to a velocity of zero, then = 4.90 × 25.0
s 2
accelerates back to the surface, just as a dropped ball would do
from the height reached. Thus, the time required to decelerate = 122.5 m

upward is the same as the time required to accelerate downward. = 120 m

The ball returns to the surface with the same velocity with which
it was thrown (neglecting friction). Th erefore, 2.19. Listing the known and unknown quantities gives
_ F = 100 N
v f – v i
a =
t m = 5 kg
at = v f – v i a = ?
v f = at + v i These are the quantities of Newton’s second law of motion,

m _
2 )
(

= 9.8 (3.0 s) F = ma, and

s F _

m _ F = ma ∴ a =
m

= (9.8)(3.0) × s
s 2 _
kg·m
100
_ 2
m·s
s
= 29 _
s·s =
5 kg
= 29 m/s
100 _ _
_ kg·m 1

= ×

2.17. These three questions are easily answered by using the three 5 s 2 kg

sets of relationships, or equations, that were presented in this m _
chapter: = 20
s 2
(a) v f = at + v i , and when v i is zero.
2.20. List the known and unknown quantities:
v f = at
m = 100 kg
m _
2)
(
v f = 9.8 (4 s) v = 6 m/s
s
m _ p = ?
= 9.8 × 4 × s

s 2 These are the quantities found in the equation for momentum,

_ p = mv, which is already solved for momentum (p). Th us,
m·s
= 39
s·s
p = mv
= 40 m/s
m _
= (100 kg) ( 6 )

− _ _ s
40 m/s + 0
=
v f + v i

(b) v = = 20 m/s

2 2 kg·m
_

= 600
s
E-5 APPENDIX E Solutions for Group A Parallel Exercises 647