Posted by **Andrea** on Monday, December 19, 2011 at 10:29am.

the speed of the train is reduce uniformly from 15m/s to 7m/s while travelling a distance of 90m.

a) compute the acceleration

b) how much further will the train travel before coming to rest provided the acceleration remain constant.

- Physics -
**drwls**, Monday, December 19, 2011 at 12:27pm
a) The average speed during the interval is (1/2)(15+7) = 11 m/s

The time required to travel 90 m while decelerating from 15 to 7 m/s is

t = 90/11 = 8.18 s

The acceleration rate during that interval is

a = (7 - 15)/8.18 = -0.978 m/s^2

b) To stop, the additional time required is (7 m/s)/(0.978 m/s^2) = 7.158 s

Multiply that by the average speed during that interval, 3.5 m/s.

25.06 m/s

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