A particle is moveint along the x-axis whose instantaneous speed is given by `v^(2)=108-9x^(2)`. The acceleration of the particle is.

Options

A. `-9x ms^(-2)`
B. `-18x ms^(-2)`
C. `(-9x)/(2) ms^(-2)`
D. None of there

Answer

A

Explanation

`v^(2)=18-9x^(2)`
`a=(dv)/(dt)=(dv)/(dx).(dx)/(dt) =(d(sqrt108-9x^(2)))/(dx).(dx)/(dt)`
`a=(1(-18x))/(2sqrt(108-9x^(2))).sqrt(108-9x^(2)) =-9x^(-2)`
Alternative: Differentiating w.r.t. we get
`2v(dv)/(dx)=-18x`
`(vdx)/(dx)=-9x rArr a=-9x because (v(dv)/(dx)=a)`.