M2 June 2012 Q1
1. [In this question \(\mathbf{i}\) and \(\mathbf{j}\) are perpendicular unit vectors in a horizontal plane.]
A particle \(P\) moves in such a way that its velocity \(\mathbf{v}\) m s\(^{-1}\) at time \(t\) seconds is given by
\[\mathbf{v} = (3t^2 - 1)\mathbf{i} + (4t - t^2)\mathbf{j}\]Given that, when \(t = 0\), the position vector of \(P\) is \(\mathbf{i}\) metres,
| Scheme | Marks |
|---|---|
| \(\mathbf{a} = \dfrac{\mathrm{d}\mathbf{v}}{\mathrm{d}t} = 6t\mathbf{i} + (4 - 2t)\mathbf{j}\) | M1 A1 |
| When \(t = 1,\ \ \mathbf{a} = 6\mathbf{i} + 2\mathbf{j}\) | DM1 |
| \(|\mathbf{a}| = \sqrt{6^2 + 2^2} = \sqrt{40} = 6.32\) (m s\(^{-2}\)) | DM1 A1 |
| (5) |
Notes
M1 Differentiate \(\mathbf{v}\) to obtain \(\mathbf{a}\).
A1 Accept column vector or \(\mathbf{i}\) and \(\mathbf{j}\) components dealt with separately.
DM1 Substitute \(t = 1\) into their \(\mathbf{a}\). Dependent on 1st M1
DM1 Use of Pythagoras to find the magnitude of their \(\mathbf{a}\). Allow with their \(t\). Dependent on 1st M1
A1 Accept awrt 6.32, 6.3 or exact equivalents.
| Scheme | Marks |
|---|---|
| \(\mathbf{r} = \displaystyle\int (3t^2 - 1)\mathbf{i} + (4t - t^2)\mathbf{j}\ \mathrm{d}t\) | M1 |
| \(= (t^3 - t + C)\mathbf{i} + (2t^2 - \tfrac{1}{3}t^3 + D)\mathbf{j}\) | A1 |
| \(t = 0,\ \mathbf{r} = \mathbf{i} \Rightarrow C = 1,\ D = 0\) | DM1 |
| When \(t = 3,\ \mathbf{r} = 25\mathbf{i} + 9\mathbf{j}\) (m) | DM1 A1 |
| (5) | |
| (10 marks) |
Notes
M1 Integrate \(\mathbf{v}\) to obtain \(\mathbf{r}\)
A1 Condone \(C\), \(D\) missing
DM1 Use \(t = 0\), \(\mathbf{r} = \mathbf{i}\) to find \(C\) & \(D\)
DM1 Substitute \(t = 3\) with their \(C\) & \(D\) to find \(\mathbf{r}\). Dependent on both previous Ms.
A1 cao. Must be a vector.