A cyclist starts from rest and moves with a constant acceleration of 1 `m//s^(2)`. A boy who is 48 m behind the cyclist starts moving with a constant velocity of `10 m//s`. After how much time the boy meets the cyclist?

Options

A. `8 s`
B. `12 s`
C. `10 s`
D. both 1 and (2)

Answer

D

Explanation

Let the boy meet the cyclist at `P`.
Boy (uniform motion): `10 t=48+d …(i)`
Cyclist (accelerated motion): `d=(1)/(2).1.t^(2)…(ii)`
`10 t=48+t^(2)/2impliest^(2)-20t+96=0`
`(t-8)(t-12)=0`
The boy crosses the cyclist
(i) `At=12 s`, velocity of cyclist, `v_(c)=1xx8=8 m//s`
`v_(b)gtv_(c)`, boy crosses the cyclist
`At=12 s`, velocity of cyclist, `v_(c)=1xx12=12 m//s`
(ii) `v_(c)gtv_(b)`, cyclist crosses the boy and will be always ahead