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

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(C) The velocity components are given by the time derivatives of position:$v_x = \frac{dx}{dt} = \frac{d}{dt}(3t^2 - 6t) = 6t - 6 	ext{ m/s}$.$v_y =

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The velocity of the particle is zero at $t = 1 	ext{ s}$.

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(C) The velocity components are given by the time derivatives of position:$v_x = \frac{dx}{dt} = \frac{d}{dt}(3t^2 - 6t) = 6t - 6 	ext{ m/s}$.$v_y = \frac{dy}{dt} = \frac{d}{dt}(t^2 - 2t) = 2t - 2 	ext{ m/s}$.At $t = 1 	ext{ s}$,$v_x = 6(1) - 6 = 0 	ext{ m/s}$ and $v_y = 2(1) - 2 = 0 	ext{ m/s}$.Since both components are zero,the magnitude of velocity $v = \sqrt{v_x^2 + v_y^2} = 0 	ext{ m/s}$ at $t = 1 	ext{ s}$.Therefore,the velocity of the particle is zero at $t = 1 	ext{ s}$.

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

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