Question

a a 10000 kg train was moving with 48m/s then starts decelerate with constant rate to stop on its stationary that is found 192m away so the train slides through this distance and stops. determine the acceleration of the train​

Answer 1

a = 6 m/s^2

Step-by-step explanation:

Suvat equation is applied

s = 192m

v (final velocity) = 0

u (initial velocity) = 48

a = ?

So,

v^2 = u^2 + 2as

(0)^2 = (48)^2 + 2a(192)

-2304 / 384 = a

a = 6 m/s^2

(negative sign is ignored as it asked as acceleration not deceleration)

I hope my answer helps you