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?

Answer 1

just use you tried and true

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

a = F/m

Answer 2

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

Answer 3

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

Answer 4

Oops. Got to use the initial velocity:

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