A hiker starts on a path at time $t=0$ and reaches destination after 1 hour. During the hike, his velocity in km/h varies according to the function $v(t) = \cos{\big(\frac{t\pi}{2}\big)}.$ Find the time in hours at which the hiker reaches the halfway distance between start and destination. Hint: $\sin{\big(\frac{\pi}{6}\big)} = \frac{1}{2}.$

How does one compute the distance travelled between two time moments given an arbitrary equation for the velocity?

… and if the time ranged between 0 and the generic time $\tau?$

You should now have a generic equation that you can solve for time, knowing the previous total distance has now halved.

The distance travelled in time $\tau$ can be calculated as follows: \begin{aligned} \int_{0}^{\tau}{\cos{\Big(\frac{t\pi}{2}\Big)}} \,dt &= \left[\frac{2}{\pi}\sin{\Big(\frac{t\pi}{2}\Big)}\right]_0^\tau \\&= \frac{2}{\pi}\sin{\Big(\frac{\tau \pi}{2}\Big)} \end{aligned} Therefore, the distance travelled in an hour is $\frac{2}{\pi}\sin{\big(\frac{\pi}{2}\big)} = \frac{2}{\pi}.$ Denoting the time in hours to travel half the distance as $x,$ we get $\frac{2}{\pi}\sin{\big(\frac{x\pi}{2}\big)} = \frac{1}{2}\cdot\frac{2}{\pi}$ which gives us $\sin{\big(\frac{x\pi}{2}\big)} = \frac{1}{2}.$ Hence, using the hint, we can deduce $x = \frac{1}{3}.$