The speed `v` of a particle moving along a straight line is given by `a+bv^(2)=x^(2)`, where `x` is its distance from the origin. The acceleration of the particle is

Options

A. `bx`
B. `x/a`
C. `x/b`
D. `x/(ab)`

Answer

C

Explanation

Given `a+bv^(2)=x^(2)`
On differentiating both sides w.r.t `t` we get
`0+b(2v(dv)/(dt))=2x(dx)/(dt)`
`impliesvb (dv)/(dt)=x (dx)/(dt)implies(dv)/(dt)=x/(vb) . (dx)/(dt)`
`implies (dv)/(dt)=x/b [ :. (dx)/(dt)=v]`