Question

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

• A. $\frac{p}{q}$
• B. $\frac{p}{\sqrt{q}}$
• C. $\frac{q}{p}$
• D. $\frac{\sqrt{q}}{p}$

Answer 1

Answer: (B)

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 , $a=\frac{dv}{dt}=\frac{dv}{dx}\frac{dx}{dt}=v\frac{dv}{dx}\left(\because v=\frac{dx}{dt}\right)$
$\therefore v=\frac{dv}{dx}=a=p-qx$
$vdv=\left(p-qx\right)dx$
Integrating both sides of the above equation, we get
$\frac{v^2}{2}=px-\frac{q{x}^2}{2}$
$v^2=2px-q{x}^2$ or $v=\sqrt{2px-q{x}^2}$
At $x=\frac{p}{q},v=v_{max}=\sqrt{2p\left(\frac{p}{q}\right)-q{\left(\frac{p}{q}\right)}^2}$
$=\frac{p}{\sqrt{q}}$