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(D) Given: Radius $r = \frac{20}{\pi} \ m$,Final velocity $v = 80 \ m/s$,Number of revolutions $n = 2$.Initial velocity $u = 0 \ m/s$.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 $v = 80 \ m/s$,Number of revolutions $n = 2$.Initial velocity $u = 0 \ m/s$.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 $\omega^2 = \omega_0^2 + 2 \alpha 	heta$:$(4 \pi)^2 = 0^2 + 2 \alpha (4 \pi)$$16 \pi^2 = 8 \pi \alpha$$\alpha = \frac{16 \pi^2}{8 \pi} = 2 \pi \ rad/s^2$.The tangential acceleration $a_t = r \alpha = \left( \frac{20}{\pi} ight) 	imes (2 \pi) = 40 \ m/s^2$.

$A$ particle moves along a circular path of radius $r = \frac{20}{\pi} \ m$ with a constant tangential acceleration. If the velocity of the particle at the end of the second revolution is $80 \ m/s$,what is the tangential acceleration?

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