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, assuming that the train speed has remained at
20.3 m/s?
Answer in units of m

V = Vo + at
T1 = (VVo)/a = (020.3)/4 = 5.075 s. to stop.
T2 = 19.4 s. @ rest.
T3 = (VVo)/a = (20.30)/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.