| Exam Board | OCR MEI |
|---|---|
| Module | Paper 1 (Paper 1) |
| Year | 2024 |
| Session | June |
| Marks | 6 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Variable acceleration (vectors) |
| Type | Cartesian equation of path |
| Difficulty | Moderate -0.5 This is a straightforward mechanics question requiring integration of velocity to find position (standard technique) and elimination of parameter t to find Cartesian equation. Both parts are routine applications of calculus with no conceptual challenges or novel problem-solving required, making it slightly easier than average. |
| Spec | 1.10h Vectors in kinematics: uniform acceleration in vector form3.02f Non-uniform acceleration: using differentiation and integration |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| \(\mathbf{r} = \int(3\mathbf{i} + (6t^2 - 5)\mathbf{j})\,dt\) | M1 | Attempt to integrate velocity either as a vector or 2 separate components |
| \(= 3t\mathbf{i} + (2t^3 - 5t)\mathbf{j} + \mathbf{c}\) | A1 | Condone missing constant |
| When \(t = 0\), \(\mathbf{r}_0 = 0\mathbf{i} + 7\mathbf{j}\) | M1 | Either as a vector constant or 2 separate components evaluated. May be implied by correct vector answer |
| Position is \(3t\mathbf{i} + (2t^3 - 5t + 7)\mathbf{j}\) | A1 | Must be in vector form. Allow \(3t\mathbf{i} + (2t^3 - 5t)\mathbf{j} + 7\mathbf{j}\) |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| Using \(x = 3t\) and \(y = 2t^3 - 5t + 7\) | M1 | Attempt to eliminate \(t\) from the parametric equations |
| \(y = 2\left(\frac{x}{3}\right)^3 - 5\left(\frac{x}{3}\right) + 7\) | A1 | FT their (a); isw |
## Question 12:
### Part (a):
| Answer | Marks | Guidance |
|--------|-------|----------|
| $\mathbf{r} = \int(3\mathbf{i} + (6t^2 - 5)\mathbf{j})\,dt$ | M1 | Attempt to integrate velocity either as a vector or 2 separate components |
| $= 3t\mathbf{i} + (2t^3 - 5t)\mathbf{j} + \mathbf{c}$ | A1 | Condone missing constant |
| When $t = 0$, $\mathbf{r}_0 = 0\mathbf{i} + 7\mathbf{j}$ | M1 | Either as a vector constant or 2 separate components evaluated. May be implied by correct vector answer |
| Position is $3t\mathbf{i} + (2t^3 - 5t + 7)\mathbf{j}$ | A1 | Must be in vector form. Allow $3t\mathbf{i} + (2t^3 - 5t)\mathbf{j} + 7\mathbf{j}$ |
### Part (b):
| Answer | Marks | Guidance |
|--------|-------|----------|
| Using $x = 3t$ and $y = 2t^3 - 5t + 7$ | M1 | Attempt to eliminate $t$ from the parametric equations |
| $y = 2\left(\frac{x}{3}\right)^3 - 5\left(\frac{x}{3}\right) + 7$ | A1 | FT their (a); isw |
---12 In this question the unit vectors $\mathbf { i }$ and $\mathbf { j }$ are in the $x$ - and $y$-directions respectively.\\
The velocity $\mathbf { v } \mathrm { m } \mathrm { s } ^ { - 1 }$ of a particle is given by $\mathbf { v } = 3 \mathbf { i } + \left( 6 t ^ { 2 } - 5 \right) \mathbf { j }$. The initial position of the particle is $7 \mathbf { j } \mathrm {~m}$.
\begin{enumerate}[label=(\alph*)]
\item Find an expression for the position vector of the particle at time $t \mathrm {~s}$.
\item Find the Cartesian equation of the path of the particle.
\end{enumerate}
\hfill \mbox{\textit{OCR MEI Paper 1 2024 Q12 [6]}}