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. 4sec
B. 8sec
C. 18sec
D. 16sec

Answer

B

Explanation

Let after a time t, the cyclist overtake the bus. Then `96 + (1)/(2) xx 2 xx t^(2) = 20 xx t or t^(2) - 20t + 96= 0`
`:. t= (20 +- sqrt(400-4 xx 96))/(2 xx 1)`
`= (20 +- 4)/(2)` = 8 sec. and 12 sec.