36. We assume the train accelerates from rest (
v
0
0
=
and
x
0
0
=
) at
a
1
2
134
=+
.m
/
s
until the midway point and then decelerates at
a
2
2
=−
.
m / s until it comes to a stop
v
2
0
=
bg
at the next station. The velocity at the midpoint is
v
1
which occurs at
x
1
= 806/2
= 403m.
(a) Eq. 216 leads to
()
22
2
10
1
1
1
2
2 1.34 m/s
403 m
vv
a
x
v
¡
=
32.9 m/s.
=
(b) The time
t
1
for the accelerating stage is (using Eq. 215)
2
1
1
1
1
2
2 403 m
1
24.53 s
21
.
3
4
m
/
s
xv
t
a
t
t
¡
==
.
Since the time interval for the decelerating stage turns out to be the same, we double this
result and obtain
t
= 49.1 s for the travel time between stations.
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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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