A street car moves rectilinearly from station $A$ to the next station $B$ (from rest to rest) with an acceleration varying according to the law $f = a - b x$ , where $a$ and $b$ are constants and $x$ is the distance from station $A$ . The distance between the two stations and the maximum velocity are

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$x = \frac{b}{2 a}, v_{\max} = \frac{a}{b}$

$x = \frac{a}{2 b}, v_{\max} = \frac{b}{\sqrt{a}}$

$x = \frac{2 a}{b}, v_{\max} = \frac{a}{\sqrt{b}}$

$x = \frac{a}{b}, v_{\max} = \frac{\sqrt{a}}{b}$

answer is A.

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Detailed Solution

$f = v \cdot \frac{d v}{d x} = a - b x$
or $\int_{0}^{v} v d v = \int_{0}^{x} (a - b x) d x$
$∴ v = \sqrt{2 a x - b x^{2}} . . . (i)$
At other station, $v = 0$
$\Rightarrow x = \frac{2 a}{b}$
Further acceleration will change its direction when

$f = 0$

or
$a - b x = 0$
or
$x = \frac{a}{b}$
At this $x$ , velocity is maximum.
Using Eq.(i),
$v_{\max} = \sqrt{2 a (\frac{a}{b}) - b {(\frac{a}{b})}^{2}} = \frac{a}{\sqrt{b}}$