Question

A train travelling at a speed of 36 km/h is brought to rest by applying brakes. if it travels 100 m before coming to rest, find the value of retardation.

Answer 1

Speed = Initial velocity = u =  36 km/h = 36 * (5/18) = 10 m/s
Distance = S = 100 m
Final velocity = v = 0 m/s
We know, 
v^2 = u^2 - 2aS (Since, retardation)
0 = (10)^2 - 2*a*100
- 100 = -200a
a = 100/200
= 0.5 m/s^2

Answer 2

Answer:

Acceleration, $a=-0.5\ m/s^2$

Explanation:

It is given that,

Initial speed of the train, u = 36 km/h = 10 m/s

Final speed of the train, v = 0 (at rest)

Distance travelled, d = 100 m

Let a is the acceleration of the train. It can be calculated using the third equation of motion as :

$v^2-u^2=2ad$

$0^2-(10)^2=2a\times 100$

$a=-0.5\ m/s^2$

So, the acceleration of the train is $0.5\ m/s^2$ and it is retarding. Hence, this is the required solution.