The position of a point at time t is given by | vedclass

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(A) Given the position coordinates: $x = a + bt - ct^2$ and $y = at + bt^2$. To find the acceleration,we first find the velocity components by differe

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$2 \sqrt{b^2 + c^2} 	ext{ unit/s}^2$

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(A) Given the position coordinates: $x = a + bt - ct^2$ and $y = at + bt^2$. To find the acceleration,we first find the velocity components by differentiating with respect to time $t$: $v_x = \frac{dx}{dt} = b - 2ct$ $v_y = \frac{dy}{dt} = a + 2bt$ Now,we find the acceleration components by differentiating velocity with respect to time $t$: $a_x = \frac{dv_x}{dt} = -2c$ $a_y = \frac{dv_y}{dt} = 2b$ The resultant acceleration $a$ is given by the magnitude of the acceleration vector: $a = \sqrt{a_x^2 + a_y^2} = \sqrt{(-2c)^2 + (2b)^2}$ $a = \sqrt{4c^2 + 4b^2} = 2 \sqrt{c^2 + b^2} 	ext{ unit/s}^2$.

The position of a point at time $t$ is given by $x = a + bt - ct^2$ and $y = at + bt^2$. The resultant acceleration of the point at time $t$ is given by:

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