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Practice Paper 3 Question 4

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: $$\sin{\big(\frac{\pi}{6}\big)} = \frac{1}{2}.$$

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Hints

• Hint 1
How does one compute the distance travelled between two time moments given an arbitrary equation for the velocity?
• Hint 2
... and if the time ranged between 0 and the generic time $$\tau?$$
• Hint 3
You should now have a generic equation that you can solve for time, knowing the previous total distance has now halved.

Solution

The distance travelled in time $$\tau$$ can be calculated as follows: \begin{align} \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{align} 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 $$x = \frac{1}{3}.$$

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