June 2023 Paper 3 Q8
8 A particle \(P\) moves with constant acceleration \((3\mathbf{i} - 2\mathbf{j})\,\mathrm{m\,s^{-2}}\). At time \(t = 4\) seconds, \(P\) has velocity \(6\mathbf{i}\,\mathrm{m\,s^{-1}}\).
Determine the speed of \(P\) at time \(t = 0\) seconds. [4]
| Scheme | Marks | AO |
|---|---|---|
| \(6\mathbf{i} = \mathbf{u} + 4(3\mathbf{i} - 2\mathbf{j})\) | M1* | 3.3 |
| \(\mathbf{u} = -6\mathbf{i} + 8\mathbf{j}\) | A1 | 2.5 |
| \(u = \sqrt{(-6)^2 + 8^2}\) | M1dep* | 1.1 |
| \(u = 10\ (\mathrm{m\,s^{-1}})\) | A1 | 1.1 |
| [4] |
Notes
M1*: Applying \(\mathbf{v} = \mathbf{u} + \mathbf{a}t\) correctly - working must imply that \(\mathbf{v}\) and \(\mathbf{a}\) are vectors
Or for \(\mathbf{v} = 3t\mathbf{i} - 2t\mathbf{j} + \mathbf{c}\) and using \(t = 4\), \(\mathbf{v} = 6\mathbf{i}\) to find \(\mathbf{c}\)
M0 if \(\mathbf{u} = -6\mathbf{i} \pm 14\mathbf{j}\)
A1: or for \(\mathbf{v} = (3t - 6)\mathbf{i} + (-2t + 8)\mathbf{j}\) and setting \(t = 0\) to obtain correct \(\mathbf{u}\)
M1dep*: Correctly taking the magnitude of their \(\mathbf{u}\) but condone \(\sqrt{-6^2 + 8^2} = \sqrt{\pm 36 + 64}\)
Correct answer following \(-6\mathbf{i} + 8\mathbf{j}\) (with no wrong working) scores full marks
A1: www