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and
a
vv vv
vx vx
=−
−
−
1
2
02 01
2
01 02
2
02 1
01 2
.
(a) Substituting
x
1
= 56.7 m,
v
01
= 80.5 km/h = 22.4 m/s,
x
2
= 24.4 m and
v
02
= 48.3 km/h
= 13.4 m/s, we find
22
02 1
01 2
01 02
02
01
(13.4 m/s) (56.7 m) (22.4 m/s) (24.4 m)
0.74 s.
(
)
(22.4 m/s)(13.4 m/s)(13.4 m/s 22.4 m/s)
r
t
vv v
v
−
−
==
=
−−
(b) In a similar manner, substituting
x
1
= 56.7 m,
v
01
= 80.5 km/h = 22.4 m/s,
x
2
= 24.4 m
and
v
02
= 48.3 km/h = 13.4 m/s gives
2
02 01
01 02
02 1
01 2
1
1 (13.4 m/s)(22.4 m/s)
(22.4 m/s)(13.4 m/s)
6.2 m/s .
2
2
(13.4 m/s)(56.7 m) (22.4 m/s)(24.4 m)
a
−
−
The
magnitude
of the deceleration is therefore 6.2 m/s
2
. Although rounded off values are
displayed in the above substitutions, what we have input into our calculators are the
“exact” values (such as
v
02
161
12
=
m/s).
85. We denote
t
r
as the reaction time and
t
b
as the braking time. The motion during
t
r
is of
the constantvelocity (call it
v
0
) type. Then the position of the car is given by
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 Spring '06
 Stoler
 Physics

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