M2 June 2014 Q2
2. At time \(t\) seconds, where \(t \geqslant 0\), a particle \(P\) is moving on a horizontal plane with acceleration \([(3t^2 - 4t)\mathbf{i} + (6t - 5)\mathbf{j}]\) m s\(^{-2}\).
When \(t = 3\) the velocity of \(P\) is \((11\mathbf{i} + 10\mathbf{j})\) m s\(^{-1}\).
Find
| Scheme | Marks |
|---|---|
| Integrate: \(\mathbf{v} = \left(t^3 - 2t^2\right)\mathbf{i} + \left(3t^2 - 5t\right)\mathbf{j} + \mathbf{C}\) | M1 A2 |
| \(t = 3:\ \mathbf{v} = 9\mathbf{i} + 12\mathbf{j} + \mathbf{C} = 11\mathbf{i} + 10\mathbf{j}\ \ \ \ \ \mathbf{C} = 2\mathbf{i} - 2\mathbf{j}\) | DM1 |
| \(\mathbf{v} = \left(t^3 - 2t^2 + 2\right)\mathbf{i} + \left(3t^2 - 5t - 2\right)\mathbf{j}\) | A1 |
| (5) |
Notes
M1 At least 3 powers going up. Condone errors in constants. Must be two separate component equations if not in vector form. Could be in column vector form. Allow with no “+ \(\mathbf{C}\)”
A2 -1 each integration error. i.e. All correct A1A1 1 error A1A0, 2 or more errors A0A0 Allow with no “+ \(\mathbf{C}\)”
DM1 Substitute given values to find \(\mathbf{C}\). Dependent on the previous M mark
A1 Correct velocity (any equivalent form)
| Scheme | Marks |
|---|---|
| Parallel to \(\mathbf{i} \Rightarrow 3t^2 - 5t - 2 = 0\) | M1 |
| \((3t + 1)(t - 2) = 0,\ \ \ \ t = 2\) | A1 |
| \(|\mathbf{v}| = 8 - 8 + 2 = 2\) (m s\(^{-1}\)) | DM1 A1 |
| (4) | |
| (9 marks) |
Notes
M1 Set \(\mathbf{j}\) component of their \(\mathbf{v}\) equal to zero and solve for \(t\) Correct answers imply method, but incorrect answers need to show method clearly.
A1 Correct only. Ignore \(-\dfrac{1}{3}\) if present.
DM1 Substitute their \(t\) to find \(\mathbf{v}\). Dependent on the previous M mark.
A1 The answer must be a scalar – the Q asks for speed. Results from negative \(t\) must be rejected.
A candidate who has no “+C” can score at most M1A2M0A0 M1A0M1A0