The CSAT and Practice[+] are designed by the Climb Foundation to help candidates. We are advocates for more opportunity to shine and less opportunity to fail, and we strive to level the playing field. 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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