A boy is moving with constant velocity `12 m//s`. When he is `32 m` behind a cyclist, the cyclist starts from rest and moves under constant acceleration `2 m//s^(2)`. After how much time the boy meets the cyclist? Explain reason for two answers.

Explanation

Let the boy meet the cyclist after time t and at distance d as shown in the figure.

Boy (uniform motion):
`32+d=vt=12t …(i)`
cyclist (aaccelerated motion):
`s=d=ut+1/2at^(2)=0+(1)/(2).2.t^(2)=t^(2) …(ii)`
Solving (i) and (ii) , we get
`32+t^(2)=12t`
`t^(2)-12t+32=0`
`t=(-(-12)+-sqrt((-12)^(2)-4xx1xx32))/(2xx1)=(12+-4)/2=(12-4)/2=4s`
`t=(12+4)/2=8s`
The boy meets the cyclist after `4 s `and `8 s`.
Explanation for two answer: At `t=4 s`, velocity of cyclist, `v=u+at=0+2xx4=8 m//s`
`v_(Cyclist)gtv_(boy)`
Hence the boy overtakes the cyclist.
At `t=8 s`, velocity of cyclist, `v=0+2xx8=16 m//s`
`v_(Cyclist)gtv_(boy)`
Hence, the cyclist overtakes the boy and will remain always ahead of the boy.