Question

A particle moves along parabolic path $x=y^2+2y+2$ in such a way that $y-$ component of velocity is constant and equal to $5\text{m/s}$ during the motion. Magnitude of acceleration of the particle

• A. $50{\text{m/s}}^2$
• B. $100{\text{m/s}}^2$
• C. $10\sqrt{2}{\text{m/s}}^2$
• D. $10{\text{m/s}}^2$

Answer 1

The correct option is A $50{\text{m/s}}^2$

Given, Path of projectile
$x=y^2+2y+2$
On differentiating both side w.r.t. time
$\frac{dx}{dt}=2y\frac{dy}{dt}+2\frac{dy}{dt}$
$\Rightarrow v_x=2v_{y}y+2v_y$ $[v_y=\frac{dy}{dt}]$
$\Rightarrow v_x=10y+10$ $[v_y=5\text{m/s},\text{given}]$
Again differentiating both side w.r.t. time,
$\frac{d{v}_x}{dt}=10\frac{dy}{dt}$
$\Rightarrow a_x=50{\text{m/s}}^2$
$[a=\frac{dv}{dt},v_y=\frac{dy}{dt}=5\text{m/s, given}]$
So, magnitude of acceleration of the particle is $50{\text{m/s}}^2$.