Q. 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 $\frac{1}{\ 3}ms^{- 1}$. The acceleration of the particle is
A) $\frac{1}{3}\widehat{j}ms^{- 2}$
B) 3$\widehat{j}ms^{- 2}$
C) $\frac{2}{3}\widehat{j}ms^{- 2}$
D) 2$\widehat{j}ms^{- 2}$
Explanation
(d) $x = 30 + 180 =$120m
$y = 9x^{2}$
$\Rightarrow \ \ \frac{dy}{dt} = v_{y} = 18x\frac{dx}{dt} = 18x\left( \frac{1}{3} \right)$
$\Rightarrow \ \ \frac{dy}{dt} = v_{y} = 6x$
$a_{y} = \frac{d^{2}y}{dt^{2}} = \frac{dv_{y}}{dt} = 6\left( \frac{dx}{dt} \right) = 6\left( \frac{1}{3} \right)$
$\Rightarrow \ a_{y} = 2ms^{- 2}\ $along $y$-axis.
Hence, $a_{y} = \left( 2ms^{- 2} \right)\widehat{j}$
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.