Question

A particle moves along a parabolic path $y=9x^2$ in such a way that the $x$ component of velocity remains constant and has a value $0.333m/s$. The magnitude of acceleration of the particle is:

• A. $1$
• B. $2$
• C. $3$
• D. $4$

Answer 1

Answer: (B)

$2$

Given, $y=9x^2$

Also, x-component of velocity $v_x=\frac{dx}{dt}=0.333m/s$

Differentiating with respect to time $t$, we get $\frac{dy}{dt}=18x\frac{dx}{dt}$

$⟹$ $v_y=18x{v}_x=18x\left(0.333\right)=5.994x$

As x component velocity is constant, so acceleration has only y-component.

So, $a=a_y=\frac{d{v}_y}{dt}=5.994\frac{dx}{dt}=5.994v_x$

$⟹$ $a=5.994\left(0.333\right)=1.996\sim 2m{s}^{-2}$