A bus starts moving with acceleration `2 ms^-2`. A cyclist `96 m` behind the bus starts simultaneously towards the bus at a constant speed of `20 m//s`. After what time will he be able to overtake the bus ?

Options

A. 8s
B. 10s
C. 12s
D. s

Answer

A

Explanation

For cycle `96+s=20xxt`
for bus `s=(1)/(2)at^(2)=(1)/(2)xx2xxt^(2)=t^(2)`
`therefore96+t^(2)=20t rArr t^(2)-20t+96=0`
`(t-8)(t-12)=0 therefore t=8` sec or 12 sec
cyclist over takes bus in 8 sec