3 problems found
A ship is sailing due west at \(V\) knots while a plane, with an airspeed of \(kV\) knots, where \(k>\sqrt{2},\) patrols so that it is always to the north west of the ship. If the wind in the area is blowing from north to south at \(V\) knots and the pilot is instructed to return to the ship every thirty minutes, how long will her outward flight last? Assume that the maximum distance of the plane from the ship during the above patrol was \(d_{w}\) miles. If the air now becomes dead calm, and the pilot's orders are maintained, show that the ratio \(d_{w}/d_{c}\) of \(d_{w}\) to the new maximum distance, \(d_{c}\) miles, of the plane from the ship is \[ \frac{k^{2}-2}{2k(k^{2}-1)}\sqrt{4k^{2}-2}. \]
The island of Gammaland is totally flat and subject to a constant wind of \(w\) kh\(^{-1},\) blowing from the West. Its southernmost shore stretches almost indefinitely, due east and west, from the coastal city of Alphabet. A novice pilot is making her first solo flight from Alphaport to the town of Betaville which lies north-east of Alphaport. Her instructor has given her the correct heading to reach Betaville, flying at the plane's recommended airspeed of \(v\) kh\(^{-1},\) where \(v>w.\) On reaching Betaport the pilot returns with the opposite heading to that of the outward flight and, so featureless is Gammaland, that she only realises her error as she crosses the coast with Alphaport nowhere in sight. Assuming that she then turns West along the coast, and that her outward flight took \(t\) hours, show that her return flight takes \[ \left(\frac{v+w}{v-w}\right)t\ \text{hours.} \] If Betaville is \(d\) kilometres from Alphaport, show that, with the correct heading, the return flight should have taken \[ t+\frac{\sqrt{2}wd}{v^{2}-w^{2}}\ \text{hours.} \]
In a certain race, runners run 5\(\,\)km in a straight line to a fixed point and then turn and run back to the starting point. A steady wind of 3\(\,\text{ms}^{-1}\) is blowing from the start to the turning point. At steady racing pace, a certain runner expends energy at a constant rate of 300\(\,\)W. Two resistive forces act. One is of constant magnitude \(50\,\text{N}\). The other, arising from air resistance, is of magnitude \(2w\,\mathrm{N}\), where \(w\,\text{ms}^{-1}\) is the runner's speed relative to the air. Give a careful argument to derive formulae from which the runner's steady speed in each half of the race may be found. Calculate, to the nearest second, the time the runner will take for the whole race. \textit{Effects due to acceleration and deceleration at the start and turn may be ignored.} The runner may use alternative tactics, expending the same total energy during the race as a whole, but applying different constant powers, \(x_{1}\,\)W in the outward trip, and \(x_{2}\,\)W on the return trip. Prove that, with the wind as above, if the outward and return speeds are \(v_{1}\,\)ms\(^{-1}\) and \(v_{2}\,\)ms\(^{-1}\) respectively, then \(v_{1}+v_{2}\) is independent of the choices for \(x_{1}\) and \(x_{2}\). Hence show that these alternative tactics allow the runner to run the whole race approximately 15 seconds faster.
Solution: Note that \(P = Fv\). Since he is running at a steady pace, we can say that \(F\) must be equal to the resistive forces (as net force is \(0\)). Therefore \(F = 50 + 2(v+3)\) on the way out. ie, \(300 = (2v + 56)v \Rightarrow 150 = v^2 + 28v \Rightarrow v = \sqrt{346}-14\) On the way back, \(F = 50 + 2(v-3)\), ie \(300 = (2v+44)v \Rightarrow 150 = v^2 +22v \Rightarrow v = \sqrt{271}-11\) Therefore the total time will be \(\frac{5000}{\sqrt{346}-150} + \frac{5000}{\sqrt{271}-11} \approx 2002\), or 33 minutes, 22 seconds. Very respectable! The total energy in this first run is \(E = Pt = 2002 \cdot 300\). Now suppose we apply two different powers as in the question, then we must have: \begin{align*} && x_1 &= 2v_1^2 + 56v_1 \\ && x_2 &= 2v_2^2 + 44v_2 \\ && E &= x_1 \frac{5000}{v_1} + x_2 \frac{5000}{v_2} \\ &&&= 5000 \left ( \frac{x_1}{v_1} + \frac{x_2}{v_2} \right) \\ \Rightarrow && \frac{x_1}{v_1} &= 2v_1 + 56 \\ && \frac{x_2}{v_2} &= 2v_2 + 44 \\ \Rightarrow && \frac{E}{5000} &= 2(v_1+v_2) + 100 \\ \Rightarrow && v_1+v_2 &\text{ is independent of the choices for }x_i \end{align*} We wish to minimize \begin{align*} && \frac{5000}{v_1} + \frac{5000}{v_2} &\underbrace{\geq}_{AM-HM} 10\,000 \cdot \frac{2}{v_1+v_2} \\ &&&= 10\,000 \cdot \frac{2}{\sqrt{346}-14+\sqrt{271}-11} \\ &&&\approx 1987 \end{align*} ie they can go 15 seconds quicker with better strategy.