| Exam Board | OCR MEI |
|---|---|
| Module | Paper 1 (Paper 1) |
| Session | Specimen |
| Marks | 9 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Variable acceleration (vectors) |
| Type | When moving parallel to given vector |
| Difficulty | Standard +0.3 This is a straightforward vector mechanics question requiring integration to find position, checking when velocity components are zero, and analyzing direction conditions. Part (a) is routine (check if |v|=0), part (b) requires setting i-component to zero (one equation), and part (c) needs integration then solving a quadratic inequality. All techniques are standard with no novel insight required, making it slightly easier than average. |
| Spec | 1.08a Fundamental theorem of calculus: integration as reverse of differentiation1.10c Magnitude and direction: of vectors1.10h Vectors in kinematics: uniform acceleration in vector form |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| Require both components zero at the same time; i component zero only when \(t = 1\) and j component only when \(t = -1\) so there are no such times | M1 | May be implied but must be clear |
| A1 | Or say j component \(\geq 2\) since \(t \geq 0\) | |
| [2] |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| Travelling due north means i component is zero and j component is \(+\)ve; need \(2t - 2 = 0\) for i component, giving \(t = 1\) | M1 | Recognise velocity vector required |
| j component \(4 > 0\) so yes at \(t = 1\) | A1 | Must test j component |
| [2] |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| Recognise use of position vector required | M1 | |
| \(\mathbf{r} = \int \mathbf{v}\,dt\) so \(\mathbf{r} = \int((2t-2)\mathbf{i} + (2t+2)\mathbf{j})dt = (t^2 - 2t + C)\mathbf{i} + (t^2 + 2t + D)\mathbf{j}\) | M1 | May use \(+C\) instead |
| \(\mathbf{r} = 3\mathbf{i} + 14\mathbf{j}\) when \(t = 3\) so \(C = 0\) and \(D = -1\); so \(\mathbf{r} = (t^2 - 2t)\mathbf{i} + (t^2 + 2t - 1)\mathbf{j}\) | A1 | |
| or \(\mathbf{a} = 2\mathbf{i} + 2\mathbf{j}\) when \(t = 3\), \(\mathbf{v} = 4\mathbf{i} + 8\mathbf{j}\); \(\mathbf{r} = (4\mathbf{i}+8\mathbf{j})(t-3) + \frac{1}{2}(2\mathbf{i}+2\mathbf{j})(t-3)^2 + 3\mathbf{i} + 14\mathbf{j}\) | M1 | Must find a but may omit \(3\mathbf{i} + 14\mathbf{j}\) |
| and so \(\mathbf{r} = (t^2 - 2t)\mathbf{i} + (t^2 + 2t - 1)\mathbf{j}\) | A1 | |
| Boat is NE of O when i and j components are equal and \(+\)ve; require \(t^2 - 2t = t^2 + 2t - 1\) so \(t = 0.25\); components \(= -0.4375\) so no | M1 | Award even if \(+\)ve not mentioned |
| A1 | Must be complete argument | |
| [5] |
## Question 12:
### Part (a):
| Answer | Marks | Guidance |
|--------|-------|----------|
| Require both components zero at the same time; **i** component zero only when $t = 1$ and **j** component only when $t = -1$ so there are no such times | M1 | May be implied but must be clear |
| | A1 | Or say **j** component $\geq 2$ since $t \geq 0$ |
| **[2]** | | |
### Part (b):
| Answer | Marks | Guidance |
|--------|-------|----------|
| Travelling due north means **i** component is zero and **j** component is $+$ve; need $2t - 2 = 0$ for **i** component, giving $t = 1$ | M1 | Recognise velocity vector required |
| **j** component $4 > 0$ so yes at $t = 1$ | A1 | Must test **j** component |
| **[2]** | | |
### Part (c):
| Answer | Marks | Guidance |
|--------|-------|----------|
| Recognise use of position vector required | M1 | |
| $\mathbf{r} = \int \mathbf{v}\,dt$ so $\mathbf{r} = \int((2t-2)\mathbf{i} + (2t+2)\mathbf{j})dt = (t^2 - 2t + C)\mathbf{i} + (t^2 + 2t + D)\mathbf{j}$ | M1 | May use $+C$ instead |
| $\mathbf{r} = 3\mathbf{i} + 14\mathbf{j}$ when $t = 3$ so $C = 0$ and $D = -1$; so $\mathbf{r} = (t^2 - 2t)\mathbf{i} + (t^2 + 2t - 1)\mathbf{j}$ | A1 | |
| **or** $\mathbf{a} = 2\mathbf{i} + 2\mathbf{j}$ when $t = 3$, $\mathbf{v} = 4\mathbf{i} + 8\mathbf{j}$; $\mathbf{r} = (4\mathbf{i}+8\mathbf{j})(t-3) + \frac{1}{2}(2\mathbf{i}+2\mathbf{j})(t-3)^2 + 3\mathbf{i} + 14\mathbf{j}$ | M1 | Must find **a** but may omit $3\mathbf{i} + 14\mathbf{j}$ |
| and so $\mathbf{r} = (t^2 - 2t)\mathbf{i} + (t^2 + 2t - 1)\mathbf{j}$ | A1 | |
| Boat is NE of O when **i** and **j** components are equal and $+$ve; require $t^2 - 2t = t^2 + 2t - 1$ so $t = 0.25$; components $= -0.4375$ so no | M1 | Award even if $+$ve not mentioned |
| | A1 | Must be complete argument |
| **[5]** | | |
---12 A model boat has velocity $\mathbf { v } = ( ( 2 t - 2 ) \mathbf { i } + ( 2 t + 2 ) \mathbf { j } ) \mathrm { m } \mathrm { s } ^ { - 1 }$ for $t \geq 0$, where $t$ is the time in seconds. $\mathbf { i }$ is the unit vector east and $\mathbf { j }$ is the unit vector north.\\
When $t = 3$, the position vector of the boat is $( 3 \mathbf { i } + 14 \mathbf { j } ) \mathrm { m }$.
\begin{enumerate}[label=(\alph*)]
\item Show that the boat is never instantaneously at rest.
\item Determine any times at which the boat is moving directly northwards.
\item Determine any times at which the boat is north-east of the origin.
\end{enumerate}
\hfill \mbox{\textit{OCR MEI Paper 1 Q12 [9]}}