A car of mass m is accelerating on a level smooth road under the action of a single F. The power delivered to the car is constant and equal to P. If the velocity of the car at an instant is v, then after travelling how much distance it becomes double?
.

Options

A. `( 7 mv ^(3))/( 3P)`
B. `( 4 m v ^(3))/( 3 P )`
C. `( mv ^(3))/( P )`
D. `( 18 mv ^(3))/( 7 P )`

Answer

A

Explanation

Instantaneous power is
`P = Fv =` constant
Now by F = ma
We get, P = (ma) v
So, `a = ( P )/(mv)………….(i)`
Here acceleration is variable, So by equation (i)
`V (dv)/(ds) = (P)/(mv)`
integrating both sides,
`int _(v) ^( 2v) v ^(2) dv = (P)/(m) int _(0) ^(S) ds`
By solving we get
`S = ( 7 mv ^(3))/( 3P)`