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. `4 s`
B. `8 s`
C. `12 s`
D. `16 s`

Answer

B

Explanation

Let time `t`. The cyclist overtakes the bus, then `96` + (distance travelled by bus in time `t`) = (distance travelled by cyclist in time `t`)
`rArr (1)/(2) xx 2 xx t^(2)+ 96 =20 xx t rArr t^(2) -20 t +96 =0`
This gives `t=8 s` or `12 s`. Hence the explanation that bus will overaake the cyclist after `12 s`.