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 $y=px-qx^{2}$ where $p$ and $q$ are positive constants. Find the velocity of the particle at the origin of coordinates.

• A. $\sqrt{\frac{a \sqrt{1 + p^{2}}}{2 q}}$
• B. $\sqrt{\frac{a \sqrt{1 + q^{2}}}{2 p}}$
• 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{ gx ^2}{2 u ^2}\left(1+\tan ^2 \theta\right) $
We find that $g = a , \tan \theta= p$ and $\frac{ a }{2 u ^2}\left(1+\tan ^2 \theta\right)= q$ $u =$ velocity of particle at origin
$ =\sqrt{\frac{ a \left(1+\tan ^2 \theta\right)}{2 q }}=\sqrt{\frac{ a \left(1+ p ^2\right)}{2 q }} $