AS October 2020 Q3
3. At time \(t = 0\), a toy electric car is at rest at a fixed point \(O\). The car then moves in a horizontal straight line so that at time \(t\) seconds \((t \gt 0)\) after leaving \(O\), the velocity of the car is \(v\ \text{m s}^{-1}\) and the acceleration of the car is modelled as \((p + qv)\ \text{m s}^{-2}\), where \(p\) and \(q\) are constants.
When \(t = 0\), the acceleration of the car is \(3\ \text{m s}^{-2}\)
When \(t = T\), the acceleration of the car is \(\dfrac{1}{2}\ \text{m s}^{-2}\) and \(v = 4\)
| Scheme | Marks | AO |
|---|---|---|
| Use of initial condition to find \(p\) | M1 | 3.1b |
| \(t = 0\), \(v = 0\), acceleration \(= 3 \Rightarrow p = 3\) | A1 | 1.1b |
| Use \(v = 4\), acceleration \(= \dfrac{1}{2}\) | M1 | 1.1b |
| \(q = -\dfrac{5}{8}\) | A1 | 1.1b |
| Use acceleration \(= \dfrac{\mathrm{d}v}{\mathrm{d}t}\) and rearrange | M1 | 1.1b |
| \(8\dfrac{\mathrm{d}v}{\mathrm{d}t} = (24 - 5v)\) * | A1* | 2.2a |
| (6) |
Notes
M1: Use initial conditions
A1: cao
M1: Use second condition
A1: cao
M1: Use appropriate derivative and rearrange
A1*: Correct given answer
| Scheme | Marks | AO |
|---|---|---|
| Separate the variables and integrate | M1 | 3.1b |
| \(8\displaystyle\int \dfrac{\mathrm{d}v}{(24 - 5v)} = \int \mathrm{d}t\) | A1 | 1.1b |
| \(-\dfrac{8}{5}\ln(24 - 5v) = t + C\) | A1 | 1.1b |
| Use \(t = 0\), \(v = 0\) to give \(C = -\dfrac{8}{5}\ln 24\) | M1 | 1.1b |
| Substitute \(v = 4\) and find and simplify \(T\) | M1 | 1.1b |
| \(T = \dfrac{8}{5}\ln 6\) | A1 | 1.1b |
| (6) | ||
| (12 marks) |
Notes
M1: Separate the variables and integrate
A1: Correct integration (\(C\) not required)
M1: Use of limits or initial conditions to find \(C\)
M1: Use \(v = 4\) to find \(T\)
A1: Correct answer (single log)