M3 June 2009 Q6
6. A cyclist and her bicycle have a combined mass of 100 kg. She is working at a constant rate of 80 W and is moving in a straight line on a horizontal road. The resistance to motion is proportional to the square of her speed. Her initial speed is 4 m s\(^{-1}\) and her maximum possible speed under these conditions is 20 m s\(^{-1}\). When she is at a distance \(x\) m from a fixed point \(O\) on the road, she is moving with speed \(v\) m s\(^{-1}\) away from \(O\).
(a) Show that \[v\dfrac{\mathrm{d}v}{\mathrm{d}x} = \dfrac{8000 - v^3}{10000v}.\] (5)
(b) Find the distance she travels as her speed increases from 4 m s\(^{-1}\) to 8 m s\(^{-1}\). (5)
(c) Use the trapezium rule, with 2 intervals, to estimate how long it takes for her speed to increase from 4 m s\(^{-1}\) to 8 m s\(^{-1}\). (4)
| Scheme | Marks |
|---|---|
| At max v, driving force = resistance Driving force \(= \dfrac{80}{v}\) | B1 |
| \(\Rightarrow \dfrac{80}{20} = k \times 20^2 \Rightarrow k = \dfrac{1}{100}\) | M1A1 |
| F = ma \(\Rightarrow 100a = \dfrac{80}{v} - kv^2\ \ \left(= \dfrac{8000 - v^3}{100v}\right)\) | M1 |
| \(\Rightarrow v\dfrac{\mathrm{d}v}{\mathrm{d}x} = \dfrac{8000 - v^3}{10000v}\) * | A1 |
| Scheme | Marks |
|---|---|
| \(\displaystyle\int_4^8 \dfrac{10000v^2}{8000 - v^3}\,\mathrm{d}v = \int_0^D 1\,\mathrm{d}x\) | M1A1 |
| \(D = \left[-\dfrac{10000}{3}\ln\left|8000 - v^3\right|\right]_4^8\) | A1 |
| \(= \left(-\dfrac{10000}{3}\ln\dfrac{7488}{7936}\right) = 193.7..... \approx 194\) m (accept 190) | M1 A1 |
| Scheme | Marks |
|---|---|
| \(\dfrac{\mathrm{d}v}{\mathrm{d}t} = \dfrac{8000 - v^3}{10000v} \Rightarrow \displaystyle\int_0^T 1\,\mathrm{d}t = \int_4^8 \dfrac{10000v}{8000 - v^3}\,\mathrm{d}v\) | M1A1 |
| \(\Rightarrow T \approx \dfrac{1}{2} \times 2 \times 10000 \times \left\{\dfrac{4}{7936} + \dfrac{2 \times 6}{7784} + \dfrac{8}{7488}\right\}\) | M1 |
| \(\Rightarrow T\ (= 31.1409....) \approx 31\) | A1 |
| (14 marks) |