A particle moves in the XY-plane with x and y | vedclass

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(D) The velocity components are given by the time derivatives of the position coordinates:$v_x = \frac{dx}{dt} = \frac{d}{dt}(5t) = 5 	ext{ m/s}$$v_y

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$3$

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(D) The velocity components are given by the time derivatives of the position coordinates:$v_x = \frac{dx}{dt} = \frac{d}{dt}(5t) = 5 	ext{ m/s}$$v_y = \frac{dy}{dt} = \frac{d}{dt}(135t - 5t^3) = 135 - 15t^2 	ext{ m/s}$The acceleration components are given by the time derivatives of the velocity components:$a_x = \frac{dv_x}{dt} = 0 	ext{ m/s}^2$$a_y = \frac{dv_y}{dt} = \frac{d}{dt}(135 - 15t^2) = -30t 	ext{ m/s}^2$Two vectors are perpendicular if their dot product is zero:$\vec{v} \cdot \vec{a} = v_x a_x + v_y a_y = 0$$(5)(0) + (135 - 15t^2)(-30t) = 0$Since $t > 0$, we have:$135 - 15t^2 = 0$$15t^2 = 135$$t^2 = 9$$t = 3 	ext{ s}$

$A$ particle moves in the $XY$-plane with $x$ and $y$ coordinates varying with time $t$ as $x(t) = 5t$ and $y(t) = 5t(27 - t^2)$. At what time $t$ (in seconds) will the direction of velocity and acceleration be perpendicular to each other?

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