Question

A particle moves in the plane xy with constant acceleration a directed along the negative $y$ -axis. The equation of motion of the particle has the form $y=px-q{x}^2$ where $p$ and $q$ are positive constants. Find the velocity of the particle at the origin of coordinates.

• A. $\sqrt{\frac{a\sqrt{1+p^2}}{2q}}$
• B. $\sqrt{\frac{a\sqrt{1+q^2}}{2p}}$
• C. $\sqrt{\frac{a\sqrt{1+q^2}}{p}}$
• D. $\sqrt{\frac{a\sqrt{1+p^2}}{q}}$

Answer 1

Answer: (A)

Comparing the given equation with the equation of a projectile motion,
$y=x\tan \theta -\frac{g{x}^2}{2u^2}\left(1+{\tan }^2\theta \right)$
We find that $g=a,\tan \theta =p$ and $\frac{a}{2u^2}\left(1+{\tan }^2\theta \right)=q$ $u=$ velocity of particle at origin
$=\sqrt{\frac{a\left(1+{\tan }^2\theta \right)}{2q}}=\sqrt{\frac{a\left(1+p^2\right)}{2q}}$