M3 June 2015 Q4
4. A vehicle of mass 900 kg moves along a straight horizontal road. At time \(t\) seconds the resultant force acting on the vehicle has magnitude \(\dfrac{63000}{kt^2}\) N, where \(k\) is a positive constant. The force acts in the direction of motion of the vehicle. At time \(t\) seconds, \(t \geqslant 1\), the speed of the vehicle is \(v\) m s\(^{-1}\) and the vehicle is a distance \(x\) metres from a fixed point \(O\) on the road. When \(t = 1\) the vehicle is at rest at \(O\) and when \(t = 4\) the speed of the vehicle is 10.5 m s\(^{-1}\).
| Scheme | Marks |
|---|---|
| \(\dfrac{63000}{kt^2} = 900\dfrac{\mathrm{d}v}{\mathrm{d}t}\) | M1 |
| \(-\dfrac{70}{kt}\ \ (+c) = v\) | DM1A1ft |
| \(t = 1\ \ v = 0 \Rightarrow c = \dfrac{70}{k}\) | M1(either) |
| \(t = 4\ \ v = 10.5 \Rightarrow -\dfrac{70}{4k} + c = 10.5\) | A1(both) |
| \(-\dfrac{70}{4k} + \dfrac{70}{k} = 10.5\) | |
| \(k = 5,\ \ c = 14\) | A1 |
| \(v = 14 - \dfrac{14}{t}\) * | A1 cso |
| (7) |
Notes
M1 Forming an equation of motion with acceleration as \(\dfrac{\mathrm{d}v}{\mathrm{d}t}\) 900 or \(m\)
DM1 Attempting the integration
A1 Correct equation. Constant of integration not needed
M1 Substituting either pair of given values
A1 Obtaining correct equations using each pair of values
A1 Obtaining correct values for \(c\) and \(k\) or use \(k = 5,\ \ c = \dfrac{70}{k}\)
A1 Substituting these values to obtain the GIVEN answer
Misread eg 6300 for 63000: M1DM1A1M1A0A0A0
Other alternative Methods: Question 4(a) by definite integration
| \(900\dfrac{\mathrm{d}v}{\mathrm{d}t} = \dfrac{63000}{kt^2}\) | M1 |
| \(\displaystyle\int_0^{10.5}\mathrm{d}v = \int_1^4 \frac{70}{kt^2}\,\mathrm{d}t\) | |
| \(\left[v\right]_0^{10.5} = \left[-\dfrac{70}{kt}\right]_1^4\) | DM1A1 Integration, limits not needed |
| \(10.5\ (-0) = -\dfrac{70}{4k} + \dfrac{70}{k}\) | M1 Substitute limits |
| \(k = 5\) | A1 Correct value |
| \(\displaystyle\int_0^v \mathrm{d}v = \int_1^t \frac{14}{t^2}\,\mathrm{d}t\) | A1 Integrate again with limits as shown |
| \(v = 14 - \dfrac{14}{t}\) * | A1 Obtain GIVEN answer |
OR:
| \(900\dfrac{\mathrm{d}v}{\mathrm{d}t} = \dfrac{63000}{kt^2}\) | M1 |
| \(\displaystyle\int_0^v \mathrm{d}v = \int_1^t \frac{70}{kt^2}\,\mathrm{d}t\) | |
| \(\left[v\right]_0^v = \left[-\dfrac{70}{kt}\right]_1^t\) | DM1A1 Integration, limits not needed |
| \(v = \dfrac{70}{k}\left[-\dfrac{1}{t}\right]_1^t = \dfrac{70}{k}\left(1 - \dfrac{1}{t}\right)\) | M1 Substitute limits and \(v = 10.5,\ t = 4\) |
| \(k = 5\) | A1 Correct value |
| \(v = \dfrac{70}{5}\left(1 - \dfrac{1}{t}\right)\) | A1 substitute |
| \(v = 14 - \dfrac{14}{t}\) * | A1 Obtain GIVEN answer |
| Scheme | Marks |
|---|---|
| \(\dfrac{14}{t} > 0 \ \Rightarrow v < 14\) or \(v\) never reaches 14 | B1 |
| (1) |
Notes
B1 Must be clear that \(v < 14\) not just never = 14 \(\dfrac{14}{t} > 0\) essential
| Scheme | Marks | ||||||||||||
|---|---|---|---|---|---|---|---|---|---|---|---|---|---|
| \(7 = 14 - \dfrac{14}{t}\) | |||||||||||||
| \(\dfrac{14}{t} = 7 \quad t = 2\) | B1 | ||||||||||||
| |||||||||||||
| \(x = \dfrac{0.25}{2}\left(0 + 2 \times 2.8 + 2 \times 4.666\ldots + 2 \times 6 + 7\right)\) | M1A1 | ||||||||||||
| \(X = 4.24175\) Accept 4.2 or 4.24 | A1 | ||||||||||||
| (4) | |||||||||||||
| (12 marks) |
Notes
B1 Showing that \(t = 2\) when \(v = 7\) Award if seen as upper limit for \(t\) in trapezium rule or values 1.25, 1.5, 1.75 seen for \(t\)
M1 Using the trapezium rule. Must have 4 intervals and values of \(t\) shown in the table.
A1 Correct numbers in the trapezium rule statement.
Values of \(v\) can be in the form \(14 - \dfrac{14}{1.25}\) etc
A1 Correct final answer. It is an estimate, so 2 or 3 sf only.