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2.Motion in Straight Line
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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
A$x=2 b / c, v_{\max }=\frac{b}{\sqrt{c}}$
B$x=\frac{c}{2 b}, v_{\max }=b / c$
C$x=\frac{b}{2 c}, v_{\max }=\frac{c}{\sqrt{a}}$
D$x=b / c, v_{\max }=\frac{\sqrt{b}}{c}$
Solution
(a)
$a=(b-c x)$
or $v \cdot \frac{d v}{d x}=(b-c x)$
$\therefore \quad v \cdot d v=(b-c x) d x$
Integrating we get, $\frac{v^2}{2}=b x-\frac{c x^2}{2}$
$\therefore \quad v^2=\left(2 b x-c x^2\right)$
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,
$v_m^2=2 b \times \frac{b}{c}-c \times \frac{b^2}{c^2}=\frac{b^2}{c}$
$v_m=\frac{b}{\sqrt{c}}$
$a=(b-c x)$
or $v \cdot \frac{d v}{d x}=(b-c x)$
$\therefore \quad v \cdot d v=(b-c x) d x$
Integrating we get, $\frac{v^2}{2}=b x-\frac{c x^2}{2}$
$\therefore \quad v^2=\left(2 b x-c x^2\right)$
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,
$v_m^2=2 b \times \frac{b}{c}-c \times \frac{b^2}{c^2}=\frac{b^2}{c}$
$v_m=\frac{b}{\sqrt{c}}$
Standard 11
Physics
