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

Options

A. `12ms^(-1)`
B. `14ms^(-1)`
C. `15ms^(-1)`
D. `16ms^(-1)`

Answer

C

Explanation

Let man will catch the after t seconds. So, he will cover distance ut. Similarly distance travelled by the bus will be `(1)/(2)at^(2)`
For the givven condition
`ut=45+(1)/(2)at^(2)=45+1.25t^(2)` [As `a=2.5 ms^(-2)`]
`rArr u=(45)/(t)+1.25t`
To find the minimum value of u
`(du)/(dt)=0 rArr (du)/(dt)=(-45)/(t^(2))+1.25=0`
so we get t = 6s, then
`u=(45)/(6)+(1.25xx6)=15 ms^(-1)`