Question

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? (8s)

Answer 1

Assume starting time as t= 0

At the time the man catches the bus , let's assume that the bus has travelled a distance of "d" meters. Then the man has travelled = d+48 m

Applying equations of motion
For man: 
time elapse (t) = (d+48)/10 secs

For bus:
$S= ut+0.5at^2 \\ d=0(t)+0.5*1*t^2 \\ d=0.5t^2$

Two equations , 2 variables (t and d). Solving them u would get a quadratic equation which gives- >
t = 8,12
Min time  = 8 secs