Tuesday

October 21, 2014

October 21, 2014

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

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 5:33pmtimes 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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