If a particle moves along a line by `S=sqrt(1+t)` then its acceleration is proportional to

Options

A. square of the velocity
B. cube of the displacement
C. cube of the velocity
D. square of the displacement

Answer

C

Explanation

We have,
`s=sqrt(1+t)`
`implies (ds)/(dt)=(1)/(2sqrt(1+t))`
`implies (ds)/(dt)=(1)/(2s)`
`implies (d^(2)s)/(dt^(2))=-(1)/(2s^(2))(ds)/(dt)=-(1)/(2s^(2))xx(1)/(2s)=-(1)/(4s^(3))`
`implies (d^(2)s)/(dt^(2))=-2xx((1)/(2s))^(3)=-2((ds)/(dt))^(3)`