Asked by Kimmy

A train has a mass of 5.27E+6 kg and is moving at 53.9 km/hr. The engineer applies the brakes, which results in a net backward force of 1.30E+6 N on the train. The brakes are held on for 25.6 s. What is the new speed of the train? How far does it travel during this period?
Answered by Steve
just use you tried and true

v = v0 + at
s = vt + 1/2 at^2

a = F/m
Answered by Kimmy
I have tried the a = F/m, and it doesn't work. Also, the 2 top equations you wrote, I'm not too sure on how to use them.. I know the 53.9 km/hr is 14.97 m/s
Answered by Steve
what do you mean F/m "doesn't work"?

a = -1.30E6/5.27E6 = -.247
v = 14.97 - .247(25.6) = 8.65
s = 8.65*25.6 - .124*25.6^2 = 140.2
Answered by Steve
Oops. Got to use the initial velocity:

s = 14.97*25.6 - .124*25.6^2 = 302.0
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