Question

A train is travelling at a speed of 90 km h⁻¹. Brakes are applied so as to produce a uniform acceleration of −0.5 m s⁻². Find how far the train will go before it is brought to rest.

Answer 1

Initial velocity of train (u) = 90 km h⁻¹

= `(90 xx 1000)/3600`

= 25 m/s

Final velocity due to rest (v) = 0

Acceleration (a) = −0.5 ms⁻²

Third equation of motion

v² = u² + 2as

0 = 252 + 2 (−0.5) × s

s = 625 m

Therefore, distance covered before coming to rest = 625 m