A car moves on a straight track from station A to the station B, with an acceleration $a = (b – cx)$, where b and c are constants and x is the distance from station A. The maximum velocity between the two stations is

$b /\sqrt{c}$

The correct answer is Option (1) → $b /\sqrt{c}$

For maximum velocity, $a = 0$ and so, $0 = b – cx$ or $x = b/c$.

Now, $v\frac{dv}{dx}=b-cx$

$\int\limits_{0}^{v}dv=\int\limits_{0}^{x}(b-cx)dx$

$\frac{v^2}{2}=bx-\frac{cx^2}{2}$

$=b×\frac{b}{c}-\frac{c(b/c)^2}{2}$

or $v=\frac{b}{\sqrt{c}}$