Question

A particle moves in the x-y plane and at a time $t$ is at the point ( ${t^2}$ , ${t^3} - 2t$ ), then which of the following is/are correct?
This question has multiple correct options
A. At $t = 0$, the particle is moving parallel to the y-axis
B. At $t = 0$, the direction of velocity and acceleration are perpendicular
C. At $t = \sqrt {\dfrac{2}{3}}$, the particle is moving parallel to the x-axis
D. At $t = 0$, the particle is at rest.

Answer 1

You can start by calculating the value of ${v_x}$ , ${v_y}$ , ${a_x}$ and ${a_y}$ for the equation of displacement given in the problem, i.e. $dx =$ ( ${t^2}$ , ${t^3} - 2t$ ). Then calculate the value of ${v_x}$ , ${v_y}$ , ${a_x}$ and ${a_y}$ on time $t = 0$ . Then calculate the value of ${v_x}$ and ${v_y}$ . Then choose the correct options.

Complete answer:
Here, we are given a particle that is moving in the x-y plane. The particle at a time $t$ is at the point ( ${t^2}$ , ${t^3} - 2t$ ).

Let the velocity of the particle in the horizontal and vertical direction be ${v_x}$ and ${v_y}$ respectively and the acceleration of the particle in the horizontal and vertical direction be ${a_x}$ and ${a_y}$ respectively.

We know that ${v_x} = \dfrac{{dx}}{{dt}} = 2t$
And ${v_y} = \dfrac{{dy}}{{dt}} = 3{t^2} - 2$
And ${a_x} = \dfrac{{d{v_x}}}{{dt}} = 2$
And ${a_y} = \dfrac{{d{v_x}}}{{dt}} = 0$
At $t = 0$
${v_x} = 2 \times 0 = 0$
${v_y} = 3 \times {\left( 0 \right)^2} - 2 = - 2$
${a_x} = 2$
${a_y} = 0$
So, at $t = 0$ the velocity and acceleration of the particle is
$v = 0i - 2j$
$a = 2i + 0j$
So, at the time $t = 0$ , the particle is moving parallel to the y-axis as the particle has velocity only in the direction of the y-axis.

At $t = 0$ , the direction of velocity and acceleration are perpendicular as the velocity of the particle is in the direction of the x-axis and acceleration is the direction of the y-axis.
At $t = \sqrt {\dfrac{2}{3}}$
${v_x} = 2\left( {\sqrt {\dfrac{2}{3}} } \right)$
${v_y} = 3{\left( {\sqrt {\dfrac{2}{3}} } \right)^2} - 2 = 0$
So, the velocity at a time $t = \sqrt {\dfrac{2}{3}}$ is
$v = 2\left( {\sqrt {\dfrac{2}{3}} } \right)i + 0j$

So, at the time $t = \sqrt {\dfrac{2}{3}}$ the particle is moving parallel to the x-axis as the particle has velocity only in the direction of the x-axis.

So, the correct answer is “Option A,B and C”.

Note:
In such type of problems where we have to choose multiple options, it is usually best to calculate the position in the x-y plane for all the time intervals given in the problem (in this case $t = 0$ and $t = \sqrt {\dfrac{2}{3}}$). This will help us to choose the correct options more easily (in this case option A, B, and C).