Question

The maximum velocity and the distance AB

Answer 1

a=p-qx

$a=v\frac{dv}{dx}$

$v\frac{dv}{dx} =p-qx$

$\frac{v^{2}}{2} = px-\frac{1}{2} qx^{2}$

$v^{2} =2px-qx^{2}$

FOR DISTANCE AB , v=0

$0=2px-qx^{2}$

$x=0,x=\frac{2p}{q}$

FOR MAXIMUM VELOCITY a=0

$at , x=p/q$

$v^{2} =\frac{2p^{2}}{q} -\frac{p^{2}}{q} =\frac{p^{2}}{q}$

$v=\frac{p}{\sqrt{q}}$