Physics - Calculus Question with Solution | TestHub

The velocity is given by $v(t)=A{\cos }^2(\omega t)$.

Since ${\cos }^2(\omega t)$ is always non-negative (its range is $[0,1]$), and $A$ is a positive constant, the velocity $v(t)$ is always non-negative. This rules out option A. The minimum value of ${\cos }^2(\omega t)$ is 0, so the minimum velocity is $A\times 0=0$. This rules out option B and D. Using the trigonometric identity ${\cos }^2(\theta )=\frac{1+\cos (2\theta )}{2}$, we can write $v(t)=A\left(\frac{1+\cos (2\omega t)}{2}\right)=\frac{A}{2}+\frac{A}{2}\cos (2\omega t)$. The average value of $\cos (2\omega t)$ over a long time (or any integer number of periods) is 0. Therefore, the average velocity over a long time is $\frac{A}{2}$.