An eastbound subway train accelerates from rest at 1.22 m/s2 along a straight tr

Question

An eastbound subway train accelerates from rest at 1.22 m/s2 along a straight track.

When the train starts, a second train is 1.75 km to the east of the first.>

The second train is westbound and moves at a constant speed ofstyle="font-size: 12px;">

23.0 m/s alongstyle="font-size: 12px;"> a parallel track.

a) How far apart are the trains when the eastbound subway train is

traveling at the same speed as the westbound train?

b) How far from the starting point of the eastbound

train are the two trains when they begin to pass?

c) With what velocity is the eastbound train

traveling when they begin to pass?

Explanation / Answer

Part A)

For the east train apply

vf^2 = vo^2 + 2ad

23^2 = (0) + 2(1.22)(d)
d = 216.8 m

The time for this is found from

vf = vo + at

23 = 0 + (1.22)(t)

t = 18.85 sec

The west train travels a distance in that time

d = vt = 23(18.85) = 433.6 m

The original distance was 1750 m, so

1750 - 216.8 - 433.6 = 1099.6 m (1.0996 km)

Part B)

d of the eastbound train will be found by d = vot + .5at^2

For the westbound d = vt.

The two d values will need to be 1750 m

vt + .5at^2 = 1750

23t + .5(1.22)(t^2) = 1750

In standard form

.61t^2 + 23t - 1750 = 0

Find the roots in the quadratic equation

Roots are 37.93 sec and -75.6 sec (we can eliminate negative time)

t = 37.93

Thus d = 0 + (.5)(1.22)(37.93)^2

d = 877.6 m

Part C)
vf^2 = v0^2 + 2ad

vf^2 = (0) + 2(1.22)(877.6)

vf = 46.3 m/s

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