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. Only once after 8.0 seconds from the start
B. Only one after 12 seconds after start
C. Twice after 8 seconds, after 12 seconds from the start
D. Never

Answer

C

Explanation

We use : `x (t) =x(0 ) + ut +1//2 a t^(2) ` in such problems which involve two moving bodies or trying to catch up with each other
At t=0
Let the initial position of boy be the origin After a time t ` x _(boy) (t) =0 +10 t `
` (##VMC_PHY_XI_WOR_BOK_01_C02_E03_045_S01##) `
` x_(bus) (t) =48 0t +1//2 (L) t^(2) `
Boy catches up with the bus when they are at same position
`rArr " "x_(boy) (t) =x_(bus) (t) rArr" "0+ 10 t = 48 +0t +(1)/(2) (L) t^(2) rArr t=8s ,12s " " ` After t=8s
Boy reaches the bus and will overtake the bus (If he doesn’t enter it) ` V_(Boy) gt V_(Bus ) `
At t=12s , the bus will overtake the boy as `v_(Bus ) gt V_(Boy )`
Analysis on Graph: Bus : Parabolic x – t graph.Boy : Straight line x – t graph
` (##VMC_PHY_XI_WOR_BOK_01_C02_E03_045_S02.png" width="80%">