Answer

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**Hint:**To find the time at which the velocity and acceleration are zero, we have to find the relation between time and velocity and acceleration respectively. In order to do that, the given equation of position must be differentiated once w.r.t time for velocity and twice for acceleration. Once we know them, we can establish a relationship with time and understand when these two quantities will be zero.

**Formula used:**

$\eqalign{

& {v_x} = \dfrac{{dx}}{{dt}},{v_y} = \dfrac{{dy}}{{dt}} \cr

& {a_x} = \dfrac{{{d^2}x}}{{d{t^2}}},{a_y} = \dfrac{{{d^2}y}}{{d{t^2}}} \cr} $

**Complete answer:**

In the question, they’ve given the equation of the position of the particle at any time t.

The velocity of the particle in the x-axis is given by

${v_x} = \dfrac{{dx}}{{dt}}$

Substituting the given equation $x = \left( {3{t^2} - 6t} \right)metres$ and differentiating, we get

$\eqalign{

& {v_x} = \dfrac{{dx}}{{dt}} \cr

& \Rightarrow {v_x} = \dfrac{{d\left( {3{t^2} - 6t} \right)}}{{dt}} \cr

& \Rightarrow {v_x} = 6t - 6 \cr} $

Therefore, the velocity of the particle along the x-axis is given by ${v_x} = 6t - 6$.

The acceleration of the particle in the x-axis is given by

$\eqalign{

& {a_x} = \dfrac{{{d^2}x}}{{d{t^2}}} \cr

& \Rightarrow {a_x} = \dfrac{d}{{dt}}\left( {\dfrac{{dx}}{{dt}}} \right) \cr

& \Rightarrow {a_x} = \dfrac{d}{{dt}}\left( {{v_x}} \right) \cr} $

Substituting the given equation ${v_x} = 6t - 6$ and differentiating, we get

$\eqalign{

& {a_x} = \dfrac{d}{{dt}}\left( {{v_x}} \right) \cr

& \Rightarrow {a_x} = \dfrac{d}{{dt}}\left( {6t - 6} \right) \cr

& \Rightarrow {a_x} = 6m{s^{ - 2}} \cr} $

So, the acceleration of the particle along the x-axis will be ${a_x} = {\text{ 6m}}{{\text{s}}^{{\text{ - 2}}}}$.

Similarly, the velocity of the particle along the y-axis is

$\eqalign{

& {v_y} = \dfrac{{dy}}{{dt}} \cr

& \Rightarrow {v_y} = \dfrac{{d\left( {{t^2} - 2t} \right)}}{{dt}} \cr

& \Rightarrow {v_y} = 2t - 2 \cr} $

And the acceleration of the particle along the y-axis will be given by

$\eqalign{

& {a_y} = \dfrac{{{d^2}y}}{{d{t^2}}} \cr

& \Rightarrow {a_y} = \dfrac{d}{{dt}}\left( {{v_y}} \right) \cr

& \Rightarrow {a_y} = \dfrac{d}{{dt}}\left( {2t - 2} \right) \cr

& \Rightarrow {a_y} = 2m{s^{ - 2}} \cr} $

Thus, the velocity and acceleration of the particle along the y-axis is given by ${v_y} = 2t - 2$ and ${a_y} = 2m{s^{ - 2}}$.

As the acceleration’s x and y-components are constant, the acceleration of the particle will never be zero.

The velocity of the particle will be zero only when both, x-component and y-component are zero, simultaneously. That is

$\eqalign{

& {v_x} = 0 \cr

& \Rightarrow 6t - 6 = 0 \cr

& \Rightarrow t = 1s \cr

& {v_y} = 0 \cr

& \Rightarrow 2t - 2 = 0 \cr

& \Rightarrow t = 1s \cr} $

Therefore, when t = 1s the velocity of the particle will be zero.

**Thus, the correct option is C.**

**Note:**

The velocity is the rate of change of displacement, i.e. the change in position with respect to time. Similarly, the acceleration is the rate of change of velocity, i.e. the change in velocity with respect to time. The time at which velocity is equal to zero must be the same in both, x and y components as the magnitude will be zero only when both are zero.

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