The x and y coordinates of a particle at any | vedclass

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(A) Given the position coordinates of the particle as functions of time $t$:$x = 5t - 2t^2$$y = 10t$To find the velocity components,we differentiate t

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$-4$

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(A) Given the position coordinates of the particle as functions of time $t$:$x = 5t - 2t^2$$y = 10t$To find the velocity components,we differentiate the position coordinates with respect to time $t$:$v_x = \frac{dx}{dt} = \frac{d}{dt}(5t - 2t^2) = 5 - 4t$$v_y = \frac{dy}{dt} = \frac{d}{dt}(10t) = 10$To find the acceleration components,we differentiate the velocity components with respect to time $t$:$a_x = \frac{dv_x}{dt} = \frac{d}{dt}(5 - 4t) = -4 \, m/s^2$$a_y = \frac{dv_y}{dt} = \frac{d}{dt}(10) = 0 \, m/s^2$The acceleration vector is $\vec{a} = a_x \hat{i} + a_y \hat{j} = -4 \hat{i} + 0 \hat{j} = -4 \hat{i} \, m/s^2$.Since the acceleration components are constant,the acceleration of the particle at any time $t$,including $t = 2 \, s$,is $-4 \, m/s^2$ in the $x$-direction.

The $x$ and $y$ coordinates of a particle at any time $t$ are given by $x = 5t - 2t^2$ and $y = 10t$ respectively,where $x$ and $y$ are in meters and $t$ is in seconds. The acceleration of the particle at $t = 2 \, s$ is . . . . . . $m/s^2$.

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