# physics

posted by john

An automobile and train move together along
parallel paths at 20.3 m/s.
The automobile then undergoes a uniform
acceleration of −4 m/s2 because of a red light
and comes to rest. It remains at rest for 19.4 s,
then accelerates back to a speed of 20.3 m/s
at a rate of 1.95 m/s2.
How far behind the train is the automobile
when it reaches the speed of 20.3 m/s, as-suming that the train speed has remained at
20.3 m/s?

1. Henry

V = Vo + at
T1 = (V-Vo)/a = (0-20.3)/-4 = 5.075 s. to stop.
T2 = 19.4 s. @ rest.
T3 = (V-Vo)/a = (20.3-0)/1.95 = 10.4 s.
to regain speed.

Time Lost = T1+T2+T3 = 5.075+19.4+10.4 = 34.875 s.

d = 20.3m/s * 34.875s = 708 m. Behind
the train.

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