Q. A particle crossing the origin of co-ordinates at time $t = 0,$ moves in the $xy -$plane with a constant acceleration $a$ in the $y -$ direction. If its equation of motion is $y = bx^{2}$ ($b$ is a constant), its velocity component in the $x -$direction is
Explanation
(b) Given, $y = bx^{2}$
$\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \frac{dy}{dt} = 2bx\frac{dx}{dt}\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ldots(i)$
$\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \frac{dy}{dt} = at\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \left( \because v_{y} = u_{y} + a_{y}t \right)$
$\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ at = 2bx\frac{dx}{at}$
$\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ at\ dt = 2bx\ dx$
Take integration of both sides
$\ \ \ \ \ \ \ \ \ \ \ \ \ \int at\ dt = \int 2bx\ dx$
$\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \frac{at^{2}}{2} = bx^{2} + c\ \ \ \ \ \ \ \ \ \ \ \ldots(ii)$
$At\ t = 0,\ x = 0,\ c\text{=}0$
Then $\frac{at^{2}}{2} = bx^{2}$
$\ \ \ \ \ \ \ \ \ \ \ \ \ x = \sqrt{\frac{at^{2}}{2b}} = \sqrt{\frac{a}{2b}}t$
$\therefore\ \ \ \ \ \ v_{x} = \frac{dx}{dt} = \sqrt{\frac{a}{2b}}$
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.