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

Question

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

Answer 1

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

Answer 2

well worked,but you should be explaining in details about you find the answers

Answer 3

convert 5000g/cm³ to kg/m³ give more details

Answer 4

Great!