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Question

(C) The velocity of the first particle is given by the derivative of its position with respect to time: $v_x = \frac{dx}{dt} = 8 - 6t$. At $t = 1 \; s

Vedclass · Accepted Answer

$580$

Vedclass · Answer

(C) The velocity of the first particle is given by the derivative of its position with respect to time: $v_x = \frac{dx}{dt} = 8 - 6t$. At $t = 1 \; s$,$v_x = 8 - 6(1) = 2 \; m/s$. Thus,$\vec{v}_1 = 2 \hat{i} \; m/s$.The velocity of the second particle is given by the derivative of its position with respect to time: $v_y = \frac{dy}{dt} = -24t^2$. At $t = 1 \; s$,$v_y = -24(1)^2 = -24 \; m/s$. Thus,$\vec{v}_2 = -24 \hat{j} \; m/s$.The velocity of the second particle relative to the first particle is $\vec{v}_{21} = \vec{v}_2 - \vec{v}_1 = -2 \hat{i} - 24 \hat{j}$.The speed of the second particle in the frame of the first is the magnitude of the relative velocity: $|\vec{v}_{21}| = \sqrt{(-2)^2 + (-24)^2} = \sqrt{4 + 576} = \sqrt{580}$.Given that the speed is $\sqrt{v}$,we have $\sqrt{v} = \sqrt{580}$,which implies $v = 580$.

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