Question

A 25,000 kilogram train is traveling down a track at 20 meters per second. A cow wanders onto the tracks 75 meters ahead of the train, causing the conductor to slam on the brakes. The rain skids to a stop. If the brakes can provide 62,500 Newtons of friction, will the conductor have enough stopping distance to avoid striking the cow?

1/2(25,000)(20M/S)^2= 500,000/62500n= 80 j

Is this correct? Thank you

Answer 1

First calculated the resultant deceleration caused by the breaks using F=MA. F being given as 62500N, and M as 25,000KG. The resultant A is -2.5M/s^2.

Then using V^2 = U^2 + 2AS we can calculate the speed of the train after a displacement (s) of 75M. We have been given initial velocity (u) of 20m/s. We only have one unknown (v) which is final velocity. Solve for v...

V^2 = 20*20 + 2(-2.5*75)

V^2 = (400 - 375)
sqrt(V^2) = sqrt(25)
v = 5m/s

Poor cow. 5m/s