A particle moves in a straight line so that `s=sqrt(t)`, then its acceleration is proportional to

Options

A. `("velocity")^3`
B. velocity
C. `("velocity")^2`
D. `("velocity")^(3//2)`

Answer

A

Explanation

Given, `s=sqrt(t)`
On differenting w.r.t., we get
`(ds)/(dt)=(1)/(2sqrt(1))rArrv=(1)/(2sqrt(t))`
Again differentiating. We get
`(dv)/(dt)=-(1)/(2.2t^(3//2))`
`rArr a=-(2)/((2sqrt(t))^3)`
`rArr a=-2v^3rArr a prop v^3` .