If a particle moves such that the displacement is proportional to the square of the velocity acquired, then it acceleration is

Options

A. proportional to `s^(2)`
B. proportional to `1/(s^(2))`
C. proportiona to `1/s`
D. a constant

Answer

D

Explanation

According to question `s pro v^(2) impliesv=ksqrt(s)`
Now, acceleration `=(dv)/(dt)=k 1/(2sqrt(s)) . (ds)/(dt)`
`=k 1/(2sqrt(s)) . k sqrt(s) [ :' v=(ds)/(dt)=k sqrt(s)]`
`=(k^(2))/2=` constant