Question

A particle moves along the parabolic path $x=y^2+2y+2$ in such a way that the $y-$ component of velocity vector remains $5{\text{ms}}^{-1}$ during the motion. The magnitude of the acceleration of the particle (in $m{s}^{-2}$) is

Answer 1

Given, $v_y=\frac{dy}{dt}=5\text{m/s}$ and $a_y=0{\text{m/s}}^2$
Given, $x=y^2+2y+2$
Differentiating the above equation w.r.t $x$, we get
$\frac{dx}{dt}=2y\frac{dy}{dt}+2\frac{dy}{dt}+0$ $v_x=2y\times 5+2\times 5$ $\{\because v_y=5\}$
$v_x=10y+10$

Differentiating velocity for acceleration
$\therefore a_x=\frac{d{v}_x}{dt}=10\times \frac{dy}{dt}=10\times 5=50{\text{m/s}}^2$

Net acceleration is
$\therefore a=\sqrt{a_x^2+a_y^2}=\sqrt{(50{)}^2+0}=50{\text{m/s}}^2$