M4 June 2014 Q2
2. A car of mass 1000 kg is moving along a straight horizontal road. The engine of the car is working at a constant rate of 25 kW. When the speed of the car is \(v\) m s\(^{-1}\), the resistance to motion has magnitude \(10v\) newtons.
| Scheme | Marks |
|---|---|
| \(\dfrac{P}{v} - 10v = ma\,;\ \dfrac{25000}{v} - 10v = 1000a\) | M1 |
| \(v = 20\), (m s\(^{-2}\)) \(\quad a = \dfrac{\dfrac{25000}{20} - 10 \times 20}{1000} = \dfrac{\dfrac{25}{2} - 2}{10}\) | DM1 |
| \(= 1.05\) (m s\(^{-2}\)) ** | A1 |
| (3) |
Notes
M1 Equation of motion
DM1 Substitute \(v = 20\)
A1 Obtain given answer correctly
| Scheme | Marks |
|---|---|
| \(v\dfrac{\mathrm{d}v}{\mathrm{d}x} = \dfrac{\dfrac{25000}{v} - 10v}{1000} = \dfrac{25000 - 10v^2}{1000v} = \dfrac{2500 - v^2}{100v}\) | M1 A1 |
| \(\displaystyle\int \frac{100v^2}{2500 - v^2}\,\mathrm{d}v = \int 1\,\mathrm{d}x \qquad \left(= 100\int -1 + \frac{2500}{2500 - v^2}\,\mathrm{d}v\right)\) | M1 |
| alt1 \(\quad \displaystyle = 100\int -1 + \frac{25}{50 - v} + \frac{25}{50 + v}\,dv\) | DM1 A1 |
| \(x(+C) = 100\left\{-v + 25\ln\left|\dfrac{50 + v}{50 - v}\right|\right\}\) | A1 |
| \(x = 100\left(-20 + 25\ln\dfrac{70}{30}\right) - 100\left(-10 + 25\ln\dfrac{60}{40}\right) = 105\) (m) | DM1 A1 |
| (8) | |
| (11 marks) |
Notes
M1 Differential equation in \(v\) and \(x\)
A1 Any equivalent form
M1 Separate the variables
DM1 Split using partial fractions
A1 Or equivalent
A1 Integration correct
DM1 Correct use of limits
A1 Or better \(\left(2500\ln\left(\dfrac{14}{9}\right) - 1000\right)\)
alt2
| \(= 100\left(-v + 50\,\text{arc}\tanh\left(\dfrac{v}{50}\right)\right)\) | DM1 A1 |
| \(x(+C) = 100\left\{-v + 25\ln\left|\dfrac{50 + v}{50 - v}\right|\right\}\) | A1 |
| \(x = 100\left(-20 + 25\ln\dfrac{70}{30}\right) - 100\left(-10 + 25\ln\dfrac{60}{40}\right) = 105\) (m) | DM1 A1 |
DM1 Use of arctanh
A1 correct
A1 Convert to log form
DM1 Correct use of limits
A1 Or better \(\left(2500\ln\left(\dfrac{14}{9}\right) - 1000\right)\)
(Corrected from the printed mark scheme: the first line of alt2 is printed as \(100\left(v - 50\,\text{arc}\tanh\left(\frac{v}{50}\right)\right)\), with the signs reversed.)
NB A correct numerical answer that does not follow from integration scores no marks.