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Question: A particle starts from point A moves along a straight line path with an acceleration given by a = p ...

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

pq\frac{p}{q}

B

pq\frac{p}{\sqrt{q}}

C

qp\frac{q}{p}

D

qp\frac{\sqrt{q}}{p}

Answer

pq\frac{p}{\sqrt{q}}

Explanation

Solution

Given: a = p – qx …..(i)

⇒\Rightarrow0 = p – qx or x=pqx = \frac{p}{q}

∴\thereforeVelocity is maximum at x=pqx = \frac{p}{q}

∵\becauseAccelerations,

a=dvdt=dvdxdxdt=vdvdx(∵v=dxdt)∴vdvdx=a=p−qxa = \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=(p−qx)dxvdv = (p - qx)dx

Integrating both sides of the above equations, we get v22=px−qx22\frac{v^{2}}{2} = px - \frac{qx^{2}}{2}

v2=2px−qx2orv=2px−qx2v^{2} = 2px - qx^{2}orv = \sqrt{2px - qx^{2}}

At x=pq,v=v2P(Pq)−q(Pq)2Pqmax⁡x = \frac{p}{q},v = {v\sqrt{2P\left( \frac{P}{q} \right) - q\left( \frac{P}{q} \right)^{2}}\frac{P}{\sqrt{q}}}_{\max}