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18. We use Eq. 2-2 (average velocity) and Eq. 2-7 (average acceleration). Regarding our coordinate choices,
the initial position of the man is taken as the origin and his direction of motion during 5 min
≤
t
≤
10 min
is taken to be the positive
x
direction. We also use the fact that ∆
x
=
v
∆
t
0
when the velocity is constant
during a time interval ∆
t
0
.
(a) Here, the entire interval considered is ∆
t
=8
−
2 = 6 min which is equivalent to 360 s, whereas the
sub-interval in which he is
moving
is only ∆
t
0
=8
−
5 = 3 min = 180 s. His position at
t
=2m
in
is
x
= 0 and his position at
t
= 8 min is
x
=
v
∆
t
0
=(2
.
2)(180) = 396 m. Therefore,
v
avg
=
396 m
−
0
360 s
=1
.
10 m
/
s
.
(b) The man is at rest at
t
= 2 min and has velocity
v
=+2
.
2 m/s at
t
= 8 min. Thus, keeping the
answer to 3 signiFcant Fgures,
a
avg
=
2
.
2m
/
s
−
0
360 s
=0
.
00611 m
/
s
2
.
(c) Now, the entire interval considered is ∆
t
=9
−
3 = 6 min (360 s again), whereas the sub-interval

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