M4 June 2013 (R) Q5
5. A van of mass 1200 kg travels along a straight horizontal road against a resistance to motion which is proportional to the speed of the van. The engine of the van is working at a constant rate of 40 kW. The van starts from rest at time \(t = 0\). At time \(t\) seconds, the speed of the van is \(v\) m s\(^{-1}\). When the speed of the van is 40 m s\(^{-1}\), the acceleration of the van is 0.3 m s\(^{-2}\).
(a) Show that \[75v\frac{\mathrm{d}v}{\mathrm{d}t} = 2500 - v^2\] (6)
(b) Find \(v\) in terms of \(t\). (6)
| Scheme | Marks |
|---|---|
| \(Fv = 40000\) | |
| \(1200\dfrac{\mathrm{d}v}{\mathrm{d}t} = \dfrac{40000}{v} - kv\) | M1 A1 |
| \(\dfrac{\mathrm{d}v}{\mathrm{d}t} = 0.3 \qquad 1200 \times 0.3 = \dfrac{40000}{40} - 40k\) | M1 |
| \(k = 16\) | A1 |
| \(1200\dfrac{\mathrm{d}v}{\mathrm{d}t} = \dfrac{40000}{v} - 16v\) | |
| \(1200v\dfrac{\mathrm{d}v}{\mathrm{d}t} = 40000 - 16v^2\) | M1 |
| \(75v\dfrac{\mathrm{d}v}{\mathrm{d}t} = 2500 - v^2\) | A1 |
| (6) |
Notes
M1 Use initial conditions to find \(k\)
A1 Given Answer
| Scheme | Marks |
|---|---|
| \(75\displaystyle\int \frac{v}{2500 - v^2}\,\mathrm{d}v = \int \mathrm{d}t\) | M1 |
| \(-\dfrac{75}{2}\ln\left(2500 - v^2\right) = t \quad (+c)\) | A1 |
| \(t = 0\ \ v = 0\ \Rightarrow -\dfrac{75}{2}\ln 2500 = c\) | M1 |
| \(-\dfrac{75}{2}\ln\left(\dfrac{2500 - v^2}{2500}\right) = t\) | A1 |
| \(\dfrac{2500 - v^2}{2500} = \mathrm{e}^{-\frac{2t}{75}} \ \to\ v^2 = 2500\left(1 - \mathrm{e}^{-\frac{2t}{75}}\right)\) | M1 |
| \(v = 50\sqrt{1 - \mathrm{e}^{-\frac{2t}{75}}}\) | A1 |
| (6) | |
| (12 marks) |
Notes
M1 Separate and attempt integration
M1 Use initial values to find \(c\)
A1 Or equivalent
M1 Find \(v\) or \(v^2\) in terms of \(t\)