Question

A train is moving with a velocity of 90 km/hr. When the brakes are applied, the acceleration is reduced to 0.5 m/s. Calculate the distance covered by the train before coming to rest.

A. 225 m
B. 250 m
C. 275 m
D. 300 m

Answer 1

Using the kinematic equation v² = u² + 2as, the train, with an initial velocity of 25 m/s and deceleration of 0.5 m/s², covers a distance of 625 m before coming to a stop. However, this result does not match any of the provided multiple-choice options.

Step-by-step explanation:

The question asks for the distance covered by a train decelerating to rest from an initial velocity of 90 km/hr with a constant deceleration of 0.5 m/s2. We will use the kinematic equation v2 = u2 + 2as, where v is the final velocity (0 m/s, since it comes to rest), u is the initial velocity, a is the acceleration, and s is the distance covered.

First, we convert the initial velocity from km/hr to m/s by multiplying by ⅓ m/s per km/hr:

90 km/hr × (⅓ m/s) / (1 km/hr) = 25 m/s

Next, we plug the values into the kinematic equation:

0 = (25 m/s)2 + 2 (−0.5 m/s2)s

We solve for s to find the distance covered before coming to rest. The negative sign for acceleration indicates deceleration.

0 = 625 − s

s = 625 m / 1

s = 625 m

However, there seems to be a miscalculation here as none of the provided options (225 m, 250 m, 275 m, 300 m) match the calculated distance of 625 m.