M1 June 2013 Q7
7. [In this question, the horizontal unit vectors \(\mathbf{i}\) and \(\mathbf{j}\) are directed due east and due north respectively.]
The velocity, \(\mathbf{v}\) m s\(^{-1}\), of a particle \(P\) at time \(t\) seconds is given by
\[\mathbf{v} = (1 - 2t)\mathbf{i} + (3t - 3)\mathbf{j}\]| Scheme | Marks |
|---|---|
| \(t = 0\) gives \(\mathbf{v} = \mathbf{i} - 3\mathbf{j}\) | B1 |
| speed \(= \sqrt{1^2 + (-3)^2}\) | M1 |
| \(= \sqrt{10} = 3.2\) or better | A1 |
| (3) |
Notes
B1 for \(\mathbf{i} - 3\mathbf{j}\).
M1 for \(\sqrt{}\) (sum of squares of cpt.s)
A1 for \(\sqrt{10}\), 3.2 or better
| Scheme | Marks |
|---|---|
| \(t = 2\) gives \(\mathbf{v} = (-3\mathbf{i} + 3\mathbf{j})\) | M1 |
| Bearing is \(315^\circ\) | A1 |
| (2) |
Notes
M1 for clear attempt to sub \(t = 2\) into given expression.
A1 for 315.
| Scheme | Marks |
|---|---|
| (i) \(1 - 2t = 0 \Rightarrow t = 0.5\) | M1 A1 |
| (ii) \(-(3t - 3) = -3(1 - 2t)\) | M1 A1 |
| Solving for \(t\) | DM1 |
| \(t = 2/3\), 0.67 or better | A1 |
| (6) | |
| (11 marks) |
Notes
(i) First M1 for \(1 - 2t = 0\).
First A1 for \(t = 0.5\).
N.B. If they offer two solutions, by equating both the \(\mathbf{i}\) and \(\mathbf{j}\) components to zero, give M0.
(ii) First M1 for \(\dfrac{1 - 2t}{3t - 3} = \pm\left(\dfrac{-1}{-3}\right)\) o.e. (Must be an equation in \(t\) only)
First A1 for a correct equation (the + sign)
Second M1, dependent on first M1, for solving for \(t\).
Second A1 for 2/3, 0.67 or better.