Question

2.9 When brakes are applied, the
96 kmh' to 48 kmh in 800 m.
move before coming to rest?
(Assuming the retardation to be
constant).

2.10 In the above problem, find the
time taken by the train to stop
after the application of brake​

Answer 1

Explanation:

From 3rd equation of motion v^2-u^2=2as from first given v=96,u=48,s=800m a=-3.84 km/h^2 from same equation now take v as 48 , u as 0 and a as got now s= 300m to stop. Or can be get by same process with changing unit of velocity to m/s will get approx 266 m.

Thanks….