June 2022 Paper 1 Q9
9 In this question, the vectors \(\mathbf{i}\) and \(\mathbf{j}\) are directed east and north respectively.
The velocity \(\mathbf{v}\ \mathrm{m\,s^{-1}}\) of a particle at time \(t\) s is given by \(\mathbf{v} = kt^2\mathbf{i} + 6t\mathbf{j}\), where \(k\) is a positive constant. The magnitude of the acceleration when \(t = 2\) is \(10\ \mathrm{m\,s^{-2}}\).
The particle is at the origin when \(t = 0\).
| Scheme | Marks | AO |
|---|---|---|
| \(\mathbf{a} = \dfrac{\mathrm{d}\mathbf{v}}{\mathrm{d}t} = 2kt\mathbf{i} + 6\mathbf{j}\) | M1 | 3.1a |
| When \(t = 2\), \(\mathbf{a} = 2 \times 2k\mathbf{i} + 6\mathbf{j}\) | M1 | 1.1b |
| \(|\mathbf{a}| = \sqrt{(4k)^2 + 6^2} = 10\) giving \(16k^2 + 36 = 100\) | M1 | 3.1a |
| So \(k = 2\) | A1 | 3.2a |
| [4] |
Notes
M1: differentiating the \(\mathbf{v}\) vector
M1: substituting \(t = 2\) into their \(\mathbf{a}\) vector
M1: Equate the magnitude of their \(\mathbf{a}\) vector to 10
A1: must choose the positive value if two values seen
| Scheme | Marks | AO |
|---|---|---|
| \(\mathbf{r} = \displaystyle\int \mathbf{v}\,\mathrm{d}t = \dfrac{kt^3}{3}\mathbf{i} + 3t^2\mathbf{j} + \mathbf{c}\) | M1 | 1.1a |
| particle at the origin when \(t = 0\) so \(\mathbf{c} = \mathbf{0}\) So \(\mathbf{r} = \dfrac{kt^3}{3}\mathbf{i} + 3t^2\mathbf{j} = \left[\dfrac{2t^3}{3}\mathbf{i} + 3t^2\mathbf{j}\right]\) | A1 | 1.1b |
| [2] |
Notes
M1: integrating with their \(k\) or general \(k\). Allow for a vector or for both components separately integrated.
A1: Condone missing \(+\mathbf{c}\) or \(+\mathbf{c}\) still in their answer
FT their \(k\) if positive or general \(k\) used
Must be in vector form
| Scheme | Marks | AO |
|---|---|---|
| Northeast when the \(\mathbf{i}\) component = \(\mathbf{j}\) component \(\dfrac{2t^3}{3} = 3t^2\) | M1 | 3.1b |
| giving \(t = 4.5\) s [\(t = 0\) rejected as the particle is at the origin] | A1 | 1.1b |
| [2] |
Notes
M1: FT their \(\mathbf{r}\)
A1: www