Question

a train is travelling at a speed of 90 km per hour breaks are applied so as to produce a uniform acceleration of _I.5m per sec square. Find how far the train will go before it is in rest

Answer 1

Given, u = 90kmph = 90 * 5/18

                                 = 25 m/s.

            a = -1.5m/s^2, v = 0.


We know that v^2 = u^2 + 2 * a * s

                         0 = (25)^2 + 2 * (-1.5) * s

                         0 = 625 - 3s

                         s = 208.3m.


Hope this helps!

Answer 2

Initial speed of the train, u = 90 km/h = 25 m/s

Final speed of the train, v = 0 (finally the train comes to rest)

Acceleration = -0.5 m s-2

According to third equation of motion:

 v 2 = u 2 + 2as

Concept Insight - Wisely choose the equation of motion out of the three, to minimize the

calculations.

(0)2 = (25)2 + 2 (-0.5) s

where, s is the distance covered by the train

s=(25)²/2(0.5) = 625.

The train will cover a distance of 625 m before it comes to rest.

PLEASE MARK IT THE BRAINLEST.