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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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  • 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.


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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