Question

A particle moves along a parabolic path $y=9x^2$ in such way that the $x-$component of velocity remains constant and has a value $\frac{1}{3}m/s$. The magnitude of acceleration of the particle is

• A. $\frac{1}{3}m/s^2$
• B. $3m/s^2$
• C. $\frac{2}{3}m/s^2$
• D. $2m/s^2$

Answer 1

The correct option is D $2m/s^2$

Given, $v_x=\frac{1}{3}m/s$
Let $x$ is distance travelled in $x$ direction and t is time.
so, $v_x=\frac{x}{t}$
$\Rightarrow$ $x=v_{x}t=\frac{1}{3}t,$
Also, $\frac{dx}{dt}=v_x=\frac{1}{3}m/s$

$\frac{dy}{dt}=18x\frac{dx}{dt}=18\left(\frac{1}{3}t\right)\frac{1}{3}$
$v_y=\frac{dy}{dt}=2tm/s$
$a_y=\frac{d{v}_y}{dt}=2m/s^2$