The position of a particle moving in the xy- | vedclass

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(C) Given position coordinates are $x = 3t^2 - 6t$ and $y = t^2 - 2t$.The velocity components are obtained by differentiating position with respect to

Vedclass · Accepted Answer

The velocity of the particle is zero at $t = 1 \, s$.

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(C) Given position coordinates are $x = 3t^2 - 6t$ and $y = t^2 - 2t$.The velocity components are obtained by differentiating position with respect to time $t$:$v_x = \frac{dx}{dt} = 6t - 6$$v_y = \frac{dy}{dt} = 2t - 2$The velocity vector is $\vec{v} = v_x \hat{i} + v_y \hat{j}$.At $t = 1 \, s$:$v_x = 6(1) - 6 = 0 \, m/s$$v_y = 2(1) - 2 = 0 \, m/s$Thus,the velocity $\vec{v} = 0 \hat{i} + 0 \hat{j} = 0$ at $t = 1 \, s$.The acceleration components are obtained by differentiating velocity with respect to time $t$:$a_x = \frac{dv_x}{dt} = 6 \, m/s^2$$a_y = \frac{dv_y}{dt} = 2 \, m/s^2$Since the acceleration is constant and non-zero,it is never zero.Therefore,the correct statement is that the velocity of the particle is zero at $t = 1 \, s$.

The position of a particle moving in the $xy-$ plane at any time $t$ is given by $x = (3t^2 - 6t) \, m$ and $y = (t^2 - 2t) \, m$. Select the correct statement about the moving particle from the following:

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