Q14 of 24 Page 31

A man is 48m behind a bus which is at rest. The bus starts accelerating at the rate of 1 m/s2, at the same time the man starts running with uniform velocity of 10 m/s. What is the minimum time in which the man catches the bus?

Given, initial velocity of bus, u = 0 and acceleration, a = 1 m/s2.


When man catches the bus, distance between them = 0.


Let at time t sec, man catches the bus first.


Distance covered by man in t sec, d1 = velocity x time


= 10 m/s x t sec


= 10t m


Distance covered by bus in t sec, d2 m


As man was 48m behind, so difference between their distance covered = 48m










So, Minimum time to catch the bus will be 8sec.


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