Question

A train travelling at a speed of 90 kmph. Brakes are applied so as to produce a uniform acceleration of ‒0.5 ms-2. Find how far the train will go before it is brought to rest?

Answer 1

Answer:

625 m

Explanation:

90 kmph = 90*5/18 = 25 m/s

v²=u²+2as

⇒ 0 = (25)²-2*(0.5)(s)

⇒ s = 625 m

 

Answer 2

Explanation:

Speed of the train = u = 90 km/h (Given)

Acceleration of the train = a = -0.5 m/s² (Given)

Final velocity, v = 0 (Since train is at rest)

According to the first equation of motion -

v = u + at

0 = 25 - 0.5t ( speed = 90 × 5/18 = 25 m/s)

25 = 0.5t

= t = 50 sec

According to the second equation of motion -

S = ut - 1/2at²

S = 25 × 50 - 1/2 × .5 × 50²

= 1250 - 1/2 × .5 × 2500

= 1250 - 625

= 625 m

Hence, the distance traveled will be 625 m and the time taken will be 50 sec.