phys11_sm_01_r

57. (a)
(b)
Time
(s)
Acceleration
(m/s [W])
Equation
!
v = ! v i + ! a !t
0 –3.0 ! v = 5.0 m/s [E] + !3.0 m "
#
$
s
1.0 –3.0 ! v = 5.0 m/s [E] + !3.0 m "
#
$
s
2.0 –3.0 ! v = 5.0 m/s [E] + !3.0 m "
#
$
s
3.0 –3.0 ! v = 5.0 m/s [E] + !3.0 m "
#
$
s
4.0 –3.0 ! v = 5.0 m/s [E] + !3.0 m "
#
$
s
5.0 –3.0 ! v = 5.0 m/s [E] + !3.0 m "
#
$
s
58. (a) Reading from the graph, at t = 3.0 s,
!
d = –2.5 m [E], so the object is –2.5 m [E].
(b) Given: t = 2.0 s; position–time graph
!
Required:
Analysis:
v inst
!
v inst is equal to the slope, m, of the
tangent to the curve at t = 2.0 s, so m = ! ! d
!t .
By placing a ruler along the curve in Figure 8 at
t = 2.0 s, I can picture the tangent. The tangent has
a rise of about –5 m [E] over a run of 5.0 s.
Solution:
m = ! ! d
!t
= !5 m [E]
5.0 s
!
v = !1 m/s [E]
inst
Velocity
(m/s [W])
Copyright © 2011 Nelson Education Ltd. Chapter 1: Motion in a Straight Line 1-9
2 [E]
2 [E]
2 [E]
2 [E]
2 [E]
2 [E]
%
&
' 0.0 s
( ) 5.0
%
&
' 1.0 s
( ) 2.0
%
&
' 2.0 s
( ) –1.0
%
&
' 3.0 s
( ) –4.0
%
&
' 4.0 s
( ) –7.0
%
&
' 5.0 s
( ) –10.0
Statement: The instantaneous velocity of the
object at 2.0 s is –1 m/s [E].
(c) Given: ! ! d = –6 m [E]; ∆t = 4.0 s
!
Required:
Analysis:
Solution:
v av
!
v av = ! ! d
!t
!
v = av ! ! d
!t
= !6 m [E]
4.0 s
!
v = !1.5 m/s [E]
av
Statement: The average velocity of the object
over the time interval from 1.0 s to 5.0 s is
–1.5 m/s [E].
59. (a) Given: vi = 160 km/h; vf = 0 m/s;
a = 11.0 m/s 2
Required: ∆t
Analysis: v = v + a !t
f i av
!t = vf ! vi aav Solution: Convert vi to metres per second:
v = !160 i km " % " 1000 m % " 1 h % " 1 min
$ '
# h
$ ' $ ' $
& # 1 km & # 60 min & # 60 s
v i = 44.44 m/s
%
'
&