A street car moves rectilinearly from station | vedclass

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Question

(a)
$a=(b-c x)$
or $v \cdot \frac{d v}{d x}=(b-c x)$
$	herefore \quad v \cdot d v=(b-c x) d x$
Integrating we get, $\frac{v^2}{2}=b x-\frac{c x^2

Vedclass · Accepted Answer

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

Vedclass · Answer

(a)
$a=(b-c x)$
or $v \cdot \frac{d v}{d x}=(b-c x)$
$	herefore \quad v \cdot d v=(b-c x) d x$
Integrating we get, $\frac{v^2}{2}=b x-\frac{c x^2}{2}$
$	herefore \quad v^2=\left(2 b x-c x^2ight)$
At other station $v=0$ or $x=\frac{2 b}{c}$
Further, $v$ is maximum, where $\frac{d v}{d x}=0$
Substituting in Eq.$(ii)$ we get,
$v_m^2=2 b 	imes \frac{b}{c}-c 	imes \frac{b^2}{c^2}=\frac{b^2}{c}$
$v_m=\frac{b}{\sqrt{c}}$

A street car moves rectilinearly from station $A$ to the next station $B$ with an acceleration varying according to the law $a=(b-c x)$, where $b$ and $c$ 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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