Asked by Andrea
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) compute the acceleration
b) how much further will the train travel before coming to rest provided the acceleration remain constant.
Answers
Answered by
drwls
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
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
Answered by
Accepted son
well worked,but you should be explaining in details about you find the answers
Answered by
Accepted son
convert 5000g/cm³ to kg/m³ give more details
Answered by
Elroie
Great!
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