M4 June 2015 Q4
4. A car of mass 900 kg is moving along a straight horizontal road with the engine of the car working at a constant rate of 22.5 kW. At time \(t\) seconds, the speed of the car is \(v\) m s\(^{-1}\) \((0 < v < 30)\) and the total resistance to the motion of the car has magnitude \(25v\) newtons.
The time taken for the car to accelerate from 10 m s\(^{-1}\) to 20 m s\(^{-1}\) is \(T\) seconds.
| Scheme | Marks |
|---|---|
| Equation of motion: \(900a = \dfrac{22500}{v} - 25v\) | M1 A1 |
| \(a = \dfrac{\dfrac{22500}{v} - 25v}{900} = \dfrac{900 - v^2}{36v}\) | A1 |
| (3) |
Notes
M1 Requires all three terms. Condone sign errors
A1 Correct unsimplified equation
A1 Obtain **Given answer** with no errors seen
| Scheme | Marks |
|---|---|
| \(\dfrac{\mathrm{d}v}{\mathrm{d}t} = \dfrac{900 - v^2}{36v}\) | B1 |
| \(\displaystyle\int \frac{36v}{900 - v^2}\,\mathrm{d}v = \int 1\,\mathrm{d}t\), | M1 |
| \(t = -18\ln\left(900 - v^2\right)(+C)\) | A1 |
| \(T = -18\ln 500 + 18\ln 800 = 18\ln\dfrac{8}{5}\) | M1 A1 |
| (5) |
Notes
B1 Differential equation in \(v\) and \(t\)
M1 Separate & integrate
M1 Use limits correctly. Dependent?
A1 Obtain **Given answer** with no errors seen
| Scheme | Marks |
|---|---|
| \(\dfrac{900 - v^2}{36v} = v\dfrac{\mathrm{d}v}{\mathrm{d}x}\) | B1 |
| \(\displaystyle\int \frac{v^2}{900 - v^2}\,\mathrm{d}v = \int \frac{1}{36}\,\mathrm{d}x\) | M1 |
| \(\displaystyle = \int \frac{900}{900 - v^2} - 1\,\mathrm{d}v = \left(\int \frac{900}{60}\left(\frac{1}{30 - v} + \frac{1}{30 + v}\right) - 1\,\mathrm{d}v\right)\) | M1 |
| \(15\ln\left|\dfrac{30 + v}{30 - v}\right| - v = \dfrac{x}{36}(+C)\) | A1 |
| \(15\ln\left(\dfrac{50}{10} \times \dfrac{20}{40}\right) - (20 - 10) = \dfrac{x}{36}\) | M1 |
| \(x = 135\) (m) \(\qquad (540\ln 2.5 - 360)\) | A1 |
| (6) | |
| (14 marks) |
Notes
B1 Differential equation in \(v\) and \(x\)
M1 Separate variables
M1 Use partial fractions or equivalent
M1 Use limits and solve for \(x\). Dependent?