A ball is spun with angular acceleration alph | vedclass

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Question

(B) Given angular acceleration $\alpha = \frac{d\omega}{dt} = 6t^2 - 2t$.Integrating with respect to $t$ to find angular velocity $\omega$:$\int_{10}^

Vedclass · Accepted Answer

$\frac{t^4}{2} - \frac{t^3}{3} + 10t + 4$

Vedclass · Answer

(B) Given angular acceleration $\alpha = \frac{d\omega}{dt} = 6t^2 - 2t$.Integrating with respect to $t$ to find angular velocity $\omega$:$\int_{10}^{\omega} d\omega = \int_{0}^{t} (6t^2 - 2t) dt$$\omega - 10 = [2t^3 - t^2]_0^t$$\omega = 2t^3 - t^2 + 10$.Now,$\omega = \frac{d	heta}{dt} = 2t^3 - t^2 + 10$.Integrating with respect to $t$ to find angular position $	heta$:$\int_{4}^{	heta} d	heta = \int_{0}^{t} (2t^3 - t^2 + 10) dt$$	heta - 4 = [\frac{2t^4}{4} - \frac{t^3}{3} + 10t]_0^t$$	heta - 4 = \frac{t^4}{2} - \frac{t^3}{3} + 10t$$	heta = \frac{t^4}{2} - \frac{t^3}{3} + 10t + 4$.

$A$ ball is spun with angular acceleration $\alpha = 6t^2 - 2t$,where $t$ is in seconds and $\alpha$ is in $rad/s^2$. At $t = 0$,the ball has an angular velocity of $10 \ rad/s$ and an angular position of $4 \ rad$. The most appropriate expression for the angular position of the ball is:

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