A particle of mass m=1 kg moves in the xy-pl | vedclass

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(D) Given,$m=1 \ kg$. Force $F(t) = 2 \sin t \hat{i} + 3 \cos t \hat{j}$.Since $F = m a$,and $m=1$,$a = F = \frac{dv}{dt}$.$v_x = \int_0^{\pi/2} 2 \si

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(D) Given,$m=1 \ kg$. Force $F(t) = 2 \sin t \hat{i} + 3 \cos t \hat{j}$.Since $F = m a$,and $m=1$,$a = F = \frac{dv}{dt}$.$v_x = \int_0^{\pi/2} 2 \sin t \ dt = [-2 \cos t]_0^{\pi/2} = 2 \ m/s$.$v_y = \int_0^{\pi/2} 3 \cos t \ dt = [3 \sin t]_0^{\pi/2} = 3 \ m/s$.Magnitude of velocity $|v| = \sqrt{2^2 + 3^2} = \sqrt{13} \ m/s$.Now,$v_x = \frac{dx}{dt} = 2 \sin t \implies x = \int_0^{\pi/2} 2 \sin t \ dt = 2 \ m$.$v_y = \frac{dy}{dt} = 3 \sin t$ is incorrect; the velocity is the integral of acceleration. Since $v_x(t) = \int_0^t 2 \sin t' dt' = 2(1 - \cos t)$ and $v_y(t) = \int_0^t 3 \cos t' dt' = 3 \sin t$.Position $x = \int_0^{\pi/2} 2(1 - \cos t) dt = 2[t - \sin t]_0^{\pi/2} = 2(\frac{\pi}{2} - 1) = \pi - 2$.Position $y = \int_0^{\pi/2} 3 \sin t dt = 3[-\cos t]_0^{\pi/2} = 3(0 - (-1)) = 3$.Magnitude of position $|r| = \sqrt{(\pi-2)^2 + 3^2} = \sqrt{\pi^2 - 4\pi + 13}$.

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