Answer
When $x=0$ and $x=l$, acceleration is at a maximum
$\frac{6hv^2}{l^2}\leq k$
Work Step by Step
$P(x)=\frac{-2h}{l^3}x^3+\frac{3h}{l^2}x^2$
$P'(x)=\frac{-6h}{l^3}x^2+\frac{6h}{l^2}x$
$P''(x)=\frac{-12h}{l^3}x+\frac{6h}{l^2}\leq k$
$\frac{dx}{dt}=v$
$\frac{dP}{dt}=\frac{dP}{dx}\times\frac{dx}{dt}=\frac{-2hv}{l^3}x^3+\frac{3hv}{l^2}x^2$
$\frac{d^2P}{dt^2}=\frac{d(P'(x))}{dt}=\frac{d(\frac{dP}{dx}\cdot\frac{dx}{dt})}{dt}\cdot\frac{dx}{dt}=\big(\frac{d^2P}{dx^2}\cdot\frac{dx}{dt}+\frac{dP}{dx}\cdot\frac{d^2x}{dt^2}\big)\cdot\frac{dx}{dt}$
$\frac{d^2x}{dt^2}=0$
$\big(\frac{d^2P}{dx^2}\cdot\frac{dx}{dt}+\frac{dP}{dx}\cdot\frac{d^2x}{dt^2}\big)\cdot\frac{dx}{dt}=\frac{-12hv^2}{l^3}x+\frac{6hv^2}{l^2}$
When $x=0$ and $x=l$, acceleration is at a maximum
$\frac{6hv^2}{l^2}\leq k$
