| Exam Board | Edexcel |
|---|---|
| Module | M2 (Mechanics 2) |
| Year | 2012 |
| Session | January |
| Marks | 10 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Variable acceleration (1D) |
| Type | Vector motion with components |
| Difficulty | Moderate -0.3 This is a straightforward M2 mechanics question requiring standard techniques: substituting t=4 into velocity for speed (using Pythagoras), differentiating velocity for acceleration, and integrating velocity to find position with given initial conditions. All steps are routine applications of calculus to vectors with no problem-solving insight required, making it slightly easier than average but not trivial due to the multi-part integration work. |
| Spec | 1.07a Derivative as gradient: of tangent to curve1.08d Evaluate definite integrals: between limits1.10a Vectors in 2D: i,j notation and column vectors1.10c Magnitude and direction: of vectors3.02a Kinematics language: position, displacement, velocity, acceleration |
| Answer | Marks |
|---|---|
| \(\text{Speed} = \sqrt{8^2 + 48^2} = \sqrt{2368} = 48.7 \text{ (ms}^{-1})\) | M1 A1 |
| Answer | Marks |
|---|---|
| \(\mathbf{a} = 2\mathbf{i} - 6t\mathbf{j}\) | M1 A1 |
| When \(t = 4\), \(\mathbf{a} = 2\mathbf{i} - 24\mathbf{j}\) (ms\(^{-2}\)) | A1 |
| Answer | Marks |
|---|---|
| \(\mathbf{r} = t^2\mathbf{i} - t^3\mathbf{j} + \mathbf{C}\) | M1 A1 |
| \(t = 1, -4\mathbf{i} + \mathbf{j} = \mathbf{i} - \mathbf{j} + \mathbf{C}\), \(\mathbf{C} = -5\mathbf{i} + 2\mathbf{j}\) | DM1 |
| \(\mathbf{r} = (t^2 - 5)\mathbf{i} + (-t^3 + 2)\mathbf{j}\) | DM1 A1 |
| When \(t = 4\), \(\mathbf{r} = (16-5)\mathbf{i} + (-64 + 2)\mathbf{j} = 11\mathbf{i} - 62\mathbf{j}\) |
## (a)
| $\text{Speed} = \sqrt{8^2 + 48^2} = \sqrt{2368} = 48.7 \text{ (ms}^{-1})$ | M1 A1 |
**Total: 2 marks**
## (b)
| $\mathbf{a} = 2\mathbf{i} - 6t\mathbf{j}$ | M1 A1 |
| When $t = 4$, $\mathbf{a} = 2\mathbf{i} - 24\mathbf{j}$ (ms$^{-2}$) | A1 |
**Total: 3 marks**
## (c)
| $\mathbf{r} = t^2\mathbf{i} - t^3\mathbf{j} + \mathbf{C}$ | M1 A1 |
| $t = 1, -4\mathbf{i} + \mathbf{j} = \mathbf{i} - \mathbf{j} + \mathbf{C}$, $\mathbf{C} = -5\mathbf{i} + 2\mathbf{j}$ | DM1 |
| $\mathbf{r} = (t^2 - 5)\mathbf{i} + (-t^3 + 2)\mathbf{j}$ | DM1 A1 |
| When $t = 4$, $\mathbf{r} = (16-5)\mathbf{i} + (-64 + 2)\mathbf{j} = 11\mathbf{i} - 62\mathbf{j}$ | |
**Total: 5 marks**
**Question 2 Total: 10 marks**
---A particle $P$ is moving in a plane. At time $t$ seconds, $P$ is moving with velocity $\mathbf{v}$ m s$^{-1}$, where $\mathbf{v} = 2t\mathbf{i} - 3t^2\mathbf{j}$.
Find
\begin{enumerate}[label=(\alph*)]
\item the speed of $P$ when $t = 4$
[2]
\item the acceleration of $P$ when $t = 4$
[3]
\end{enumerate}
Given that $P$ is at the point with position vector $(-4\mathbf{i} + \mathbf{j})$ m when $t = 1$,
\begin{enumerate}[label=(\alph*)]
\setcounter{enumi}{2}
\item find the position vector of $P$ when $t = 4$
[5]
\end{enumerate}
\hfill \mbox{\textit{Edexcel M2 2012 Q2 [10]}}