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Q. A particle starts from point $A$ moves along a straight line path with an acceleration given by $a = p - qx$ where $p$, $q$ are constants and $x$ is distance from point $A$. The particle stops at point $B$. The maximum velocity of the particle is

Motion in a Straight Line

Solution:

Given : $a = p - qx$
At maximum velocity, $a = 0$
$\Rightarrow 0=p-qx$ or $x=\frac{p}{q}$
$\therefore $ Velocity is maximum at $x=\frac{p}{q}$
$\because$ Acceleration , $a=\frac{dv}{dt}=\frac{dv}{dx} \frac{dx}{dt}=v \frac{dv}{dx}\quad\left(\because v=\frac{dx}{dt}\right)$
$\therefore v=\frac{dv}{dx}=a=p-qx$
$vdv=\left(p-qx\right)dx$
Integrating both sides of the above equation, we get
$\frac{v^{2}}{2}=px-\frac{qx^{2}}{2}$
$v^{2}=2\,px-qx^{2}$ or $v=\sqrt{2\,px-qx^{2}}$
At $x=\frac{p}{q}, v=v_{max}=\sqrt{2p\left(\frac{p}{q}\right)-q\left(\frac{p}{q}\right)^{2}}$
$=\frac{p}{\sqrt{q}}$