(a) The entire interval considered is
Δ
t
= 8 – 2 = 6 min which is equivalent to 360 s,
whereas the subinterval in which he is
moving
is only
8
5
3min
180 s.
t'
Δ=−=
=
His
position at
t
= 2 min is
x
= 0 and his position at
t
= 8 min is
x vt
==
Δ
'
(2.2)(180)
396 m
=
. Therefore,
v
avg
m
s
m/s
=
−
=
396
0
360
110
..
(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 significant figures,
a
avg
2
s
=
−
=
22
0
360
0 00611
.
(c) Now, the entire interval considered is
Δ
t
= 9 – 3 = 6 min (360 s again), whereas the
subinterval in which he is moving is
Δ
t
'm
i
n
=−=
=
9 5
4
240 s ). His position at
3 min
t
=
is
x
= 0 and his position at
t
= 9 min is
xv
t
=
Δ
'(
.
)
( )
2 2 240
528 m .
Therefore,
v
avg
m
s
m/s.
=
−
=
528
0
360
147
.
(d) The man is at rest at
t
= 3 min and has velocity
v
= +2.2 m/s at
t
= 9 min.
Consequently,
a
avg
= 2.2/360 = 0.00611 m/s
2
just as in part (b).
(e) The horizontal line near the bottom of this
x
vs
t
graph represents the man standing at
x
= 0 for 0
≤
t
< 300 s and the linearly rising line for 300
≤
t
≤
600 s represents his
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This note was uploaded on 09/03/2009 for the course PHYS 1200 taught by Professor Stoler during the Spring '06 term at Rensselaer Polytechnic Institute.
 Spring '06
 Stoler
 Physics, Acceleration

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