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29. We separate the motion into two parts, and take the direction of motion to be positive.
In part 1, the vehicle accelerates from rest to its highest speed; we are given v
0
= 0; v =
20 m/s and
a
= 2.0 m/s
2
.
In part 2, the vehicle decelerates from its highest speed to a
halt; we are given v
0
= 20 m/s; v = 0 and
a
= –1.0 m/s
2
(negative because the acceleration
vector points opposite to the direction of motion).
(a) From Table 21, we find
t
1
(the duration of part 1) from v = v
0
+
at
. In this way,
1
20 0 2.0
t
=+
yields
t
1
= 10 s.
We obtain the duration
t
2
of part 2 from the same
equation. Thus, 0 = 20 + (–1.0)
t
2
leads to
t
2
= 20 s, and the total is
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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

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