A man is 45m behind the bus when the bus...

A man is 45m behind the bus when the bus start accelerating from rest with acceleration `2.5 ms^(-2)`. With what minimum velocity should the man start running to catch the bus

A

12 m/s

B

14 m/s

C

15 m/s

D

16 m/s

Text Solution

Verified by Experts

The correct Answer is:

C

Let man will batch the bus after `t` sec. So he will cover distance Similarly distance travelled by the bus will be `(1)/(2)at^2`. For the given condition
`ut=45+(1)/(2)at^2=45+(1)/(25)t^2`
`implies=u=(45)/(t)+1.25t`
To find the minimum value of u
`(du)/(dt)=0` so we get `t=6` sec, then
`u=(45)/(6)+1.25xx6=7.5=15(m)/(s)`

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