Question

A train of mass 6.84 x 10^6 kg is moving at a speed of 80 km/h. The brakes, which
produce a net backward force of 1.93 x 10^6N, are applied for 25.0 seconds.
(a) What is the new speed of the train?
(b) How far has the train travelled in this time?​

Answer 1

Explanation:

Dynamics

given

mass of train =6.84×10^6 kg

initial velocity of train =80 km/h= 22.2m/s

approx

Then breaks is applied for 25 seconds means retardation of train.

force of break =1.93×10^6 N

retardation is going to be ,

we know that a=F/m

therefore a= -0.28 m/s^2 approx negative sign means retardation

initial velocity u = 22.2 m/a

time t =25 s

retardation a = -0.28 m/s^2

to find final velocity v using first equation of motion

v = u + at

v = 22.2 + -0.28 . 0.25

v = 15.2 m/s

hence the final velocity of the train is15.2 m/s

by second equation s = ut + 1/2at^

s = 22.2×25 + 1/2 × - 0.28(0.25)^2

s = 467.5 m

hence it covers 467.5 m in 25 m after applying brake