Velocity at a specific time

Since velocity is the derivative of position, $v(t) = s'(t) = \dfrac{d}{dt}[\tan(t)] = \sec^2(t)$. This uses the standard derivative rule for tangent.

Substituting $t = \pi/4$ gives $v(\pi/4) = \sec^2(\pi/4) = (\sqrt{2})^2 = 2$. So the particle is moving in the positive direction at a rate of 2 ft/s at that instant.

As $t \to \pi/2^-$, $\cos(t) \to 0^+$, so $\sec(t) \to \infty$ and therefore $v(t) = \sec^2(t) \to \infty$. The velocity grows without bound as $t$ approaches $\pi/2$ from the left.

This corresponds to the vertical asymptote of $s(t) = \tan(t)$ at $t = \pi/2$: the particle's position increases without bound in finite time, meaning it moves infinitely far in the positive direction as $t$ approaches $\pi/2$, with its speed increasing without bound as well.