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31. We choose the positive direction to be that of the initial velocity of the car (implying that
a<
0 since
it is slowing down). We assume the acceleration is constant and use Table 2-1.
(a) Substituting
v
0
= 137 km
/
h=38
.
1m
/
s,
v
=90km
/
h=25m
/
s, and
a
=
−
5
.
2m
/
s
2
into
v
=
v
0
+
at
,
we obtain
t
=
25 m
/
s
−
38 m
/
s
−
5
.
/
s
2
=2
.
5s
.
(b) Wetakethecartobeat
x
= 0 when the brakes are applied
(at time
t
= 0). Thus, the
coordinate of the car as a
function of time is given by
x
= (38)
t
+
1
2
(
−
5
.
2)
t
2
in SI units. This function

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