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

Question

Dr. Abubacker · Accepted Answer

(b) $y = bx^{2}$
Differentiating w.r.t. $t,$ on both sides, we get
$\frac{dy}{dt} = b2x\frac{dx}{dt} \Rightarrow v_{y} = 2bxv_{x}$
Again, differentiating w.r.t. $t,$ on both sides, we get
$\frac{dv_{y}}{dt} = 2bv_{x}\frac{dx}{dt} + 2bx\frac{dv_{x}}{dt} = 2bv_{x}^{2} + 0$
$\left\lbrack \frac{dv_{x}}{dt} \right.\  = 0$, because the particle has constant acceleration along $y$-direction]
As per question
$\frac{dv_{y}}{dt} = a = 2bv_{x}^{2};\ v_{x}^{2} = \frac{a}{2b} \Rightarrow v_{x} = \sqrt{\frac{a}{2b}}$

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

Reviewed by Dr. Abubacker

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