Question

a train is moving with a velocity of 90km/h
It is brought to stop by applying the brakes
which produce a retardation of 0.5 m/sFind:
1) velocity after 30s, and
2)the time taken by the
train to come to rest.​

Answer 1

Answer:

Equation to be applied  " v = u - a×t ", v is final velocity, u is initial velocity, a is retardation and t is time.

Velocity at the time of applying break = 90 km/h  = 90 km/h × (5/18) = 25 m/s

(1) velocity after 30 s :-   v = 25-(1/2)×30 = 19 m/s

(2) time to come to rest :-  0 = 25-(1/2)×t  , solving for t ,  we get  t = 50 s

Answer 2

Solution:

A)

initial velocity (u) = 90km/h = 90*5/18

= 20m/s

final velocity (v) = ?

time taken (t) = 30s

acceleration (a) = -0.5m/s^2 ( negative symbol tells that it is retardation)

v = u + at

v = 20 + 0.5*30

v = 20 + 15

v = 35m/s