AS June 2019 Q2
2. A car moves in a straight line along a horizontal road. The car is modelled as a particle.
At time \(t\) seconds, where \(t \geqslant 0\), the speed of the car is \(v\ \text{m s}^{-1}\)
At the instant when \(t = 0\), the car passes through the point \(A\) with speed \(2\ \text{m s}^{-1}\)
The acceleration, \(a\ \text{m s}^{-2}\), of the car is modelled by
\[a = \frac{4}{2+v}\]in the direction of motion of the car.
At the instant when the car passes through the point \(B\), the speed of the car is \(4\ \text{m s}^{-1}\)
| Scheme | Marks | AO |
|---|---|---|
| \(a = \dfrac{4}{2+v} \Rightarrow \displaystyle\int (2+v)\,\mathrm{d}v = \int 4\,\mathrm{d}t\) | M1 | 2.1 |
| \(\dfrac{(2+v)^2}{2} = 4t + C_1\) | M1 A1 | 1.1b 1.1b |
| \(t = 0,\ v = 2 \Rightarrow C_1 = 8\) | M1 | 3.4 |
| \(\dfrac{(2+v)^2}{2} = 4t + 8\) | A1 | 1.1b |
| \((2+v)^2 = 8t + 16,\quad v = \sqrt{8t+16} - 2\) * | A1* | 2.2a |
| (6) |
Notes
M1: Form differential equation in \(v\) and \(t\) and prepare to integrate.
M1: Integrate to obtain \(k(2+v)^2\) or equivalent
A1: Correct integration. Condone missing constant of integration.
M1: Use the model to find the value of constant of integration.
A1: Correct solution in any form
A1*: Obtain given solution from correct working. Allow use of quadratic formula.
Alternative (a)
| Scheme | Marks | AO |
|---|---|---|
| \(a = \dfrac{4}{2+v} \Rightarrow \displaystyle\int (2+v)\,\mathrm{d}v = \int 4\,\mathrm{d}t\) | M1 | 2.1 |
| \(2v + \dfrac{v^2}{2} = 4t + C_2\) | M1 A1 | 1.1b 1.1b |
| \(t = 0,\ v = 2 \Rightarrow C_2 = 6\) | M1 | 3.4 |
| \(2v + \dfrac{v^2}{2} = 4t + 6\) | A1 | 1.1b |
| \(4v + v^2 = 8t + 12,\quad (v+2)^2 = 8t + 16\) \(\Rightarrow v = \sqrt{8t+16} - 2\) * | A1* | 2.2a |
| (6) |
Notes for the alternative:
M1: Form differential equation in \(v\) and \(t\) and prepare to integrate.
M1: Integrate to obtain \(k(2+v)^2\) or equivalent
A1: Correct integration. Condone missing constant of integration.
M1: Use the model to find the value of constant of integration.
A1: Correct solution in any form
A1*: Obtain given solution from correct working.
| Scheme | Marks | AO |
|---|---|---|
| \(v = 4 \Rightarrow 36 = 8t + 16 \Rightarrow t = 2.5\) | B1 | 1.1b |
| \(\dfrac{\mathrm{d}x}{\mathrm{d}t} = \sqrt{8t+16} - 2\) | M1 | 3.3 |
| \(x = k(8t+16)^{\frac{3}{2}} - 2t + C\) | M1 | 1.1b |
| \(x = \dfrac{1}{12}(8t+16)^{\frac{3}{2}} - 2t + C\) | A1 | 1.1b |
| \(t = 0,\ x = 0 \Rightarrow \dfrac{64}{12} + C = 0,\ C = -\dfrac{16}{3}\) | M1 | 3.4 |
| \(AB = \dfrac{1}{12}(36)^{\frac{3}{2}} - 5 - \dfrac{16}{3} = \dfrac{23}{3}\ \text{(m)}\) | A1 | 1.1b |
| (6) | ||
| (12 marks) |
Notes
B1: Use the result from (a) to find \(t\) when \(v = 4\): seen or implied
M1: Form differential equation in \(x\) and \(t\)
M1: Integrate to obtain terms of the correct form. Condone missing constant of integration.
A1: Correct integration. Condone missing constant of integration.
M1: Use boundary conditions in the model to find constant of integration, or as limits on a definite integral.
Note this is an independent M mark.
M0 if they use \(t = 4\)
A1: Correct answer only. 7.7 (m) or better