Homework Help: physics

Posted by Swag on Tuesday, September 27, 2011 at 4:20pm.

An automobile and train move together along
parallel paths at 39.1 m/s.
The automobile then undergoes a uniform
acceleration of −4 m/s
2
because of a red lightand comes to rest. It remains at rest for 22.5 s,
then accelerates back to a speed of 39.1 m/s
at a rate of 2.53 m/s
2
.
How far behind the train is the automobile
when it reaches the speed of 39.1 m/s, assuming that the train speed has remained at
39.1 m/s?
Answer in units of m

physics - bobpursley, Tuesday, September 27, 2011 at 5:33pm
times involved:
deacceleratin: 39.1/4 seconds
sittn still: 22 seconds
accelerating: 38.1/2.53 seconds

distance traveled: time deacc*39.1/2 + timeaccelerating*39.1/2

avg velocity= distance traveled/total time.

distance behind train: avgvelocitydifference*timetotalelapsed
=(39.1-avgvelocity)*totaltimeelpased.

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