A bus of 6 m long starts moving with an acceleration of 2ms−2. A cyclist 90 m behind the bus starts simultaneously towards the bus at 20 ms−1. After what minimum time will he be able to overtake the bus?
For boy to overtake,
In t seconds bus travels a distance of x meters and boy travels (x+90+6) meters
Using second equation of motion for bus,
x=ut+at22,
x=at22=2×t22=t2...........................(1)
For boy, Distance=speed×time
x+96=vt,
x+96=20t .........................................(2)
using (1) and (2),
t2−20t+96=0
we get t=12 and 8,
Since minimum time,
Hence, boy will catch the bus at t=8 seconds.