Question

The position of a particle moving in the x-y plane at any time $t$ is given by : $x=(3t^3-6t)$ metres; $y=(t^2-2t)$ metres. Select the correct statement.

• A. Acceleration is zero at $t=0$
• B. Velocity is zero at $t=0$
• C. Velocity is zero at $t=1s$
• D. Velocity and acceleration of the particle are never zero

Answer 1

The correct option is D Velocity and acceleration of the particle are never zero

Given : $x=(3t^3-6t)$ metres and $y=(t^2-2t)$ metres

For x direction : $v_x=\frac{dx}{dt}=9t^2-6$

$a_x=\frac{d^{2}x}{d{t}^2}=18t$

For y direction : $v_y=\frac{dy}{dt}=2t-2$

$a_y=\frac{d^{2}x}{d{t}^2}=2$

Hence at $t=0$, $a_x=0$

But $a_y=2=constant$, thus acceleration can never become zero

Also, $v_y=0$ at $t=1$ s but $v_x\ne 0$ at $t=1$ s

And at $t=\sqrt{\frac{2}{3}}$ s, $v_x=0$ but $v_y\ne 0$

Thus $v_x$ and $v_y$ can never become zero at the same instant. So, velocity an acceleration can never become zero.