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(D) Given: Radius $r = \frac{20}{\pi} \; m$,final velocity $v = 80 \; m/s$,initial velocity $u = 0$,and number of revolutions $n = 2$.The total angula

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$40 \; m/s^2$

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(D) Given: Radius $r = \frac{20}{\pi} \; m$,final velocity $v = 80 \; m/s$,initial velocity $u = 0$,and number of revolutions $n = 2$.The total angular displacement $	heta = 2\pi 	imes n = 2\pi 	imes 2 = 4\pi \; rad$.The final angular velocity $\omega = \frac{v}{r} = \frac{80}{20/\pi} = 4\pi \; rad/s$.Using the rotational kinematic equation $\omega^2 = \omega_0^2 + 2\alpha	heta$,where $\omega_0 = 0$:$(4\pi)^2 = 0 + 2 	imes \alpha 	imes 4\pi$$16\pi^2 = 8\pi\alpha$$\alpha = \frac{16\pi^2}{8\pi} = 2\pi \; rad/s^2$.The tangential acceleration $a_t$ is given by $a_t = r\alpha$.$a_t = \left(\frac{20}{\pi}ight) 	imes 2\pi = 40 \; m/s^2$.

$A$ particle moves along a circle of radius $\frac{20}{\pi} \; m$ with constant tangential acceleration. If the velocity of the particle is $80 \; m/s$ at the end of the second revolution after motion has begun,the tangential acceleration is:

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