Question

A particle moves along a parabolic path $y=-9x^2+6x$ in such a way that the x- component of velocity remains constant and has a value $\frac{1}{3}$ $m/s$, then the y-component of velocity of the particle at $t=1s$ is

• A. $\frac{1}{3}$ $m/s$
• B. $3$ $m/s$
• C. $\frac{2}{3}$ $m/s$
• D. $0$ $m/s$

Answer 1

The correct option is D $0$ $m/s$

$\frac{dy}{dt}=-18x\frac{dx}{dt}+6\frac{dx}{dt}$

$x=v.t=\frac{1}{3}t,\frac{dx}{dt}=v_x=\frac{1}{3}m/s\frac{dy}{dt}=-18x\frac{dx}{dt}+6\times \frac{1}{3}=-18\frac{1}{3}\left(\frac{1}{3}t\right)+2v_y=\frac{dy}{dt}=-2t+2$

At, $t=1s$
$v_y=-2\times 1+2=0m/s$