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

The correct option is B $\frac{p}{\sqrt{q}}$

Given = $a=P-qx$

$V\frac{dv}{dx}=P-qx$

${\int }_0^{v}vdv={\int }_0^x(P-qx)dx$

$\frac{v^2}{2}=Px-q\frac{x^2}{2}$ ...(1)

because body come to fast , hence velocity will be maximum at point where a

= 0 or $\frac{dv}{dx}=0$ , $P-qx=0$ For Vmax

$x=\frac{P}{q}$

$V^{2}max=2P\left(\frac{P}{q}\right)$

$V_{max}^2=2P\left(\frac{P}{q}\right)-q{\left(\frac{P}{q}\right)}^2=\frac{P^2}{q}$

$V_{max}=\sqrt{\frac{P^2}{q}}=\frac{P}{\sqrt{q}}$

Hence (B) is correct answer