If the velocity of a particle is given by `v=(180-16x)^((1)/(2))(m)/(s)`, then its acceleration will be

Options

A. Zero
B. `8 "m/s"^2`
C. `-8 "m/s"^2`
D. `4 "m/s"^2`

Answer

C

Explanation

`v=(180-16x)^(1//2)`
As `a =(dv)/(dt) = (dv)/(dx) . (dx) /(dt)`
`therefore a=1/2 (180-16x)^(-1//2) xx (16) ((dx)/(dt))`
`=-8(180-16x)^(-1//2) xx v`
`=-8(180-16x)^(-1//2) xx (180-16x)^(1//2)=-8 m//s^2`