Question

A particle moves in x-y plane and at time $t$ is at the point $(t^2,t^3-2t),$ then which of the following is/are correct?

• A. At t = 0, particle is moving parallel to y-axis
• B. At t = 0, direction of velocity and acceleration are perpendicular
• C. At $t=\sqrt{\frac{2}{3}},$particle is moving parallel to x-axis.
• D. At t = 0, particle is at rest.

Answer 1

Answer: (A), (B), (C)

The correct options are
A At t = 0, particle is moving parallel to y-axis
B At t = 0, direction of velocity and acceleration are perpendicular
C At $t=\sqrt{\frac{2}{3}},$particle is moving parallel to x-axis.

at t = 0 $v_x=2\left(0\right)=0$ $y=3(0{)}^2-2=-2$

hence parallel to y -axis

$v_x=\frac{dx}{dt}=2t=0$ $v_y=\frac{dy}{dt}=3t^2-2=-2$

$a_x=\frac{d^{2}x}{d{t}^2}=2$ $a_y=\frac{d^{2}y}{d{t}^2}=0$

therefore at t= 0

$v=0i-2j$

$a=2i+0j$

at $t=\sqrt{2/3}$

$v_x=2\left(\sqrt{2/3}\right)$ $v_y=3(\sqrt{2/3}{)}^2-2=0$

hence moving parallel to x axis