2 problems found
A train starts from a station. The tractive force exerted by the engine is at first constant and equal to \(F\). However, after the speed attains the value \(u\), the engine works at constant rate \(P,\) where \(P=Fu.\) The mass of the engine and the train together is \(M.\) Forces opposing motion may be neglected. Show that the engine will attain a speed \(v\), with \(v\geqslant u,\) after a time \[ t=\frac{M}{2P}\left(u^{2}+v^{2}\right). \] Show also that it will have travelled a distance \[ \frac{M}{6P}(2v^{3}+u^{3}) \] in this time.
Solution: While the force is constant, the train is accelerating at \(\frac{F}{M}\), and since \(u = \frac{F}{M} t_1 \Rightarrow t_1 = \frac{Mu^2}{Fu} = \frac{Mu^2}{P}\). Once the train is being driven at a constant rate, we can observe that change in energy will be power times time, ie \(Pt_2 = \frac{1}{2}M(v^2 - u^2) \Rightarrow t_2 = \frac{M}{2P} ( v^2 - u^2)\). Therefore the total time will be \(t_1 + t_2 = \frac{M}{2P} ( u^2 + v^2)\). During the first period, the distance will be: \(s_1 = \frac12 \frac{F}{M} t_1^2 = \frac12 \frac{F}{M} \frac{M^2u^2}{F^2} = \frac{Mu^3}{2P}\) In the second period, \(P = Fu\) and so \(\text{Force} = \frac{P}{v} \Rightarrow M v \frac{\d v}{\d x} = \frac{P}{v} \Rightarrow M \l \frac{v^3}{3} - \frac{u^3}{3}\r = Ps_2\) and therefore total distance will be: \(\frac{M}{6P}(2v^{3}+u^{3})\)
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.