A train was moving at the rate of 60 kmph when brakes were applied. It came to rest within a distance of 425 m. Calculate the retardation produced in the train.

The final velocity of the particle v=0

The initial velocity of the particle, u= 60 kmph

Let's multiply u with 5/18 to convert it to ms^-1

$u=60\times \frac{5}{18} m{s}^{-1}$

$=16.67{\mathrm{ms}}^{-1}$

s=425 m

Retardation is always against the velocity of the particle.

v² =u² +2as

0 = (16.67)² + 2a (425)

0 = 277.88 + a x 850

a = -0.32 ms^-2

Hence, retardation = 0.32 ms^-2