posted by Andrea on .
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.
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.