| Exam Board | Edexcel |
|---|---|
| Module | M2 (Mechanics 2) |
| Year | 2012 |
| Session | June |
| Marks | 10 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Variable acceleration (vectors) |
| Type | Magnitude of acceleration at given time |
| Difficulty | Moderate -0.5 This is a straightforward mechanics question requiring standard differentiation of velocity to find acceleration, then integration of velocity to find position. Both operations involve simple polynomial terms with no conceptual challenges—slightly easier than average due to routine application of calculus techniques. |
| Spec | 1.10h Vectors in kinematics: uniform acceleration in vector form3.02f Non-uniform acceleration: using differentiation and integration3.02g Two-dimensional variable acceleration |
| Answer | Marks | Guidance |
|---|---|---|
| \(a = \frac{dv}{dt} = 6\mathbf{i} + (4-2t)\mathbf{j}\) | M1 | Differentiate \(v\) to obtain \(a\) |
| When \(t=1\), \(a = 6\mathbf{i} + 2\mathbf{j}\) | A1 | Accept column vector or \(\mathbf{i}\) and \(\mathbf{j}\) components dealt with separately |
| \(\ | \mathbf{a}\ | = \sqrt{6^2 + 2^2} = \sqrt{40} = 6.32 \text{ (m s}^{-2})\) |
| DM1 | Use of Pythagoras to find the magnitude of their \(a\). Allow with their \(t\). Dependent on 1st M1 | |
| A1 | Accept \(\text{awrt } 6.32, 6.3\) or exact equivalents | |
| (5) |
| Answer | Marks | Guidance |
|---|---|---|
| \(\mathbf{r} = \int (3t^2 - 1)\mathbf{i} + (4t - t^2)\mathbf{j} \, dt\) | M1 | Integrate \(v\) to obtain \(\mathbf{r}\) |
| \(= (t^3 - t + C)\mathbf{i} + (2t^2 - \frac{1}{3}t^3 + D)\mathbf{j}\) | A1 | Condone \(C, D\) missing |
| When \(t=0\), \(\mathbf{r} = \mathbf{i} \Rightarrow C=1, D=0\) | DM1 | Use \(t=0, \mathbf{r}=\mathbf{i}\) to find \(C \& D\) |
| When \(t=3\), \(\mathbf{r} = 25\mathbf{i} + 9\mathbf{j}\) (m) | DM1 | Substitute \(t=3\) with their \(C \& D\) to find \(\mathbf{r}\). Dependent on both previous M's |
| A1 | cao. Must be a vector | |
| (5) | ||
| 10 |
**(a)**
| $a = \frac{dv}{dt} = 6\mathbf{i} + (4-2t)\mathbf{j}$ | M1 | Differentiate $v$ to obtain $a$ |
| When $t=1$, $a = 6\mathbf{i} + 2\mathbf{j}$ | A1 | Accept column vector or $\mathbf{i}$ and $\mathbf{j}$ components dealt with separately |
| $\|\mathbf{a}\| = \sqrt{6^2 + 2^2} = \sqrt{40} = 6.32 \text{ (m s}^{-2})$ | DM1 | Substitute $t=1$ into their $a$. Dependent on 1st M1 |
| | DM1 | Use of Pythagoras to find the magnitude of their $a$. Allow with their $t$. Dependent on 1st M1 |
| | A1 | Accept $\text{awrt } 6.32, 6.3$ or exact equivalents |
| | (5) | |
**(b)**
| $\mathbf{r} = \int (3t^2 - 1)\mathbf{i} + (4t - t^2)\mathbf{j} \, dt$ | M1 | Integrate $v$ to obtain $\mathbf{r}$ |
| $= (t^3 - t + C)\mathbf{i} + (2t^2 - \frac{1}{3}t^3 + D)\mathbf{j}$ | A1 | Condone $C, D$ missing |
| When $t=0$, $\mathbf{r} = \mathbf{i} \Rightarrow C=1, D=0$ | DM1 | Use $t=0, \mathbf{r}=\mathbf{i}$ to find $C \& D$ |
| When $t=3$, $\mathbf{r} = 25\mathbf{i} + 9\mathbf{j}$ (m) | DM1 | Substitute $t=3$ with their $C \& D$ to find $\mathbf{r}$. Dependent on both previous M's |
| | A1 | cao. Must be a vector |
| | (5) |
| | 10 | |
---\begin{enumerate}
\item \hspace{0pt} [In this question $\mathbf { i }$ and $\mathbf { j }$ are perpendicular unit vectors in a horizontal plane.]
\end{enumerate}
A particle $P$ moves in such a way that its velocity $\mathbf { v } \mathrm { m } \mathrm { s } ^ { - 1 }$ at time $t$ seconds is given by
$$\mathbf { v } = \left( 3 t ^ { 2 } - 1 \right) \mathbf { i } + \left( 4 t - t ^ { 2 } \right) \mathbf { j }$$
(a) Find the magnitude of the acceleration of $P$ when $t = 1$
Given that, when $t = 0$, the position vector of $P$ is i metres,\\
(b) find the position vector of $P$ when $t = 3$\\
\hfill \mbox{\textit{Edexcel M2 2012 Q1 [10]}}