A student is standing at a distance of 50 metre from the bus. As soon as the bus begins its motion with an acceleration of `1m//s^(2)`, the student starts running towards the bus with a uniform velocity u. assuming the motion to be along a straight road, the minimum value of u. so that the student is able to catch the bus is

Options

A. `8ms^(-1)`
B. `5ms^(-1)`
C. `12ms^(-1)`
D. `10ms^(-1)`

Answer

D

Explanation

`s=u+(1)/(2)alphat^(2)=0+(1)/(2)xx1xxt^(2)=(t^(2))/(2)`

Suppose the student catches the bus at C. then,
`(50+(t^(2))/(2))=ut` . . (i)
`impliesu=(50)/(t)+(t)/(2)` . . (ii)
To find minimum value of u, `(du)/(dt)=0`
`(du)/(dt)=(txx0-50xx1)/(t^(2))+(2xx1-1x0)/(4)`
`=-(50)/(t^(2))+(1)/(2)=0 implies tpm10s`
`therefore`From equation (i) putting, t=10s,
we get u=10 m/s.