M4 January 2005 Q4
4. A car of mass \(M\) moves along a straight horizontal road. The total resistance to motion of the car is modelled as having constant magnitude \(R\). The engine of the car works at a constant rate \(RU\).
Find the time taken for the car to accelerate from a speed of \(\tfrac{1}{4}U\) to a speed of \(\tfrac{1}{2}U\). (9)
| Scheme |
|---|
| Apply \(F = ma\): \(\ m\dfrac{\mathrm{d}v}{\mathrm{d}t} = \dfrac{RU}{v} - R\) |
| Separate the variables: \(\ \displaystyle\int_{\frac{U}{4}}^{\frac{U}{2}} \frac{mv\,\mathrm{d}v}{R(U - v)} = \int_0^T \mathrm{d}t\). |
| \(\displaystyle\frac{m}{R}\int_{\frac{U}{4}}^{\frac{U}{2}} \left(-1 + \frac{U}{(U - v)}\right)\mathrm{d}v = [t]_0^T = T\) |
| \(\dfrac{m}{R}\Big[-v - U\ln\left|U - v\right|\Big]_{\frac{U}{4}}^{\frac{U}{2}} = T\) |
| \(T = \dfrac{m}{R}\left(\left(-\tfrac{1}{2}U - U\ln\left(\tfrac{1}{2}U\right)\right) - \left(-\tfrac{1}{4}U - U\ln\left(\tfrac{3}{4}U\right)\right)\right)\) |
| \(T = \dfrac{mU}{R}\left(-\tfrac{1}{4} + \ln\left(\tfrac{3}{2}\right)\right)\) |
Notes
The published mark scheme for this paper is a set of worked answers: no mark allocation is printed.
The answers use \(m\) for the mass of the car, which the question calls \(M\).
(Corrected from the printed mark scheme: the limits of integration are printed as \(\tfrac{u}{4}\) and \(\tfrac{u}{2}\).)