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

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(D) The position coordinates are given by $x = a + bt - ct^2$ and $y = at + bt^2$.First,we find the velocity components by differentiating with respec

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$2\sqrt{b^2 + c^2}$

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(D) The position coordinates are given by $x = a + bt - ct^2$ and $y = at + bt^2$.First,we find the velocity components by differentiating with respect to $t$:$v_x = \frac{dx}{dt} = b - 2ct$$v_y = \frac{dy}{dt} = a + 2bt$Next,we find the acceleration components by differentiating the velocity components with respect to $t$:$a_x = \frac{dv_x}{dt} = \frac{d^2x}{dt^2} = -2c$$a_y = \frac{dv_y}{dt} = \frac{d^2y}{dt^2} = 2b$The resultant acceleration $a$ is given by the magnitude of the acceleration vector:$a = \sqrt{a_x^2 + a_y^2}$$a = \sqrt{(-2c)^2 + (2b)^2}$$a = \sqrt{4c^2 + 4b^2}$$a = 2\sqrt{b^2 + c^2}$

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

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