Velocity and acceleration at a given time

Differentiate $s(t) = \cos(t) + \sin(t)$ to get $v(t) = s'(t) = -\sin(t) + \cos(t) = \cos(t) - \sin(t)$. Differentiate again to get $a(t) = v'(t) = -\cos(t) - \sin(t)$.

Substitute $t = \pi/4$: since $\sin(\pi/4) = \cos(\pi/4) = \frac{\sqrt{2}}{2}$, we get $v(\pi/4) = \frac{\sqrt{2}}{2} - \frac{\sqrt{2}}{2} = 0$. Similarly, $a(\pi/4) = -\frac{\sqrt{2}}{2} - \frac{\sqrt{2}}{2} = -\sqrt{2}$.

The velocity is $v(t) = \cos(t) - \sin(t)$, which can be written as $v(t) = \sqrt{2}\cos\left(t + \frac{\pi}{4}\right)$. Since cosine ranges between $-1$ and $1$, $v(t)$ ranges between $-\sqrt{2}$ and $\sqrt{2}$.

The maximum value of $|v(t)|$ is therefore $\sqrt{2}$, achieved whenever $\cos\left(t + \frac{\pi}{4}\right) = \pm 1$, which occurs at points within $[0, 2\pi]$ such as $t = 3\pi/4$ and $t = 7\pi/4$. Thus the maximum speed of the particle on this interval is $\sqrt{2}$ meters per second.

At time pi over four, the velocity was found to be zero, so the speed there is $|v(\pi/4)| = 0$. Since the maximum speed on the interval is $\sqrt{2}$, we compare: the speed at $t = \pi/4$ is $0$, while speeds elsewhere on $[0, 2\pi]$ range up to $\sqrt{2}$. Therefore, the speed at $t = \pi/4$ represents the global minimum speed on the interval $[0, 2\pi]$. The answer is minimum.