A point moves in a straight line under the retardation `av^(2)`, where `'a'` is a positive constant and `v` is speed. If the initial speed is `u`, the distance covered in `'t'` seconds is `:`

Options

A. `a u t`
B. `(1)/(a) log(a u t)`
C. `(1)/(a) log(1 + a u t)`
D. `a log(a u t)`

Answer

C

Explanation

The retardation is given by `(dv)/(dt) = -av_(2)`
integrating between proper limits
`rArr - underset(u) overset(v)(int) (dv)/(v^(2))=underset(0) overset(t) (int) a dt`
or `(1)/(v)= at +(1)/(u)`
`rArr (dt)/(dx)=at+(1)/(u) rArr dx =(u dt)/(1+aut)`
integrating between proper limits
`rArr underset(0)overset(v) (int) (dv)/(v^(2))= underset(0)overset(t)(int) a dt`
or `(1)/(v)=at+(1)/(u)`
`rArr (dt)/(dx)=at+(1)/(u) rArr dx =(u dt)/(1+aut)`
integrating between proper limits
`rArr underset(0) overset(s) (int) dx= underset(0) overset(t)(int) (udt)/(1+aut)`
`rArr S=(1)/(a)ln (1+aut)`.