| Exam Board | Edexcel |
|---|---|
| Module | M2 (Mechanics 2) |
| Marks | 10 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Variable acceleration (1D) |
| Type | Position vector from velocity integration |
| Difficulty | Moderate -0.3 This is a straightforward M2 mechanics question requiring differentiation of velocity to find acceleration, then integration of velocity with initial conditions to find displacement. Part (a) involves basic differentiation and magnitude calculation; part (b) is a 'show that' requiring integration and substitution at t=6. All techniques are standard with no novel problem-solving required, making it slightly easier than average. |
| Spec | 1.10h Vectors in kinematics: uniform acceleration in vector form3.02g Two-dimensional variable acceleration |
| Answer | Marks | Guidance |
|---|---|---|
| (a) \(a = \frac{dv}{dt} = 3\mathbf{i} - 2\mathbf{j}\) and when \(t = 2\), \(a = 3\mathbf{i} - 4\mathbf{j}\) | M1 A1 | |
| mag. of a = \(\sqrt{(3)^2 + (-4)^2} = 5\) ms\(^{-2}\) | M1 A1 | |
| (b) \(s = \int v \, dt = \frac{3}{2}t^2 - \frac{1}{4}t^4\mathbf{j} + A\mathbf{i} + B\mathbf{j}\) | M1 A1 | |
| when \(t = 0\), \(s = 6\mathbf{i} + 12\mathbf{j}\) so \(A = 6, B = 12\) | M1 | |
| \(s = (\frac{3}{2}t^2 + 6)\mathbf{i} + (12 - \frac{1}{4}t^4)\mathbf{j}\) | A1 | |
| disp. when \(t = 6\) is \(60\mathbf{i} - 60\mathbf{j} = 60(\mathbf{i} - \mathbf{j}) \therefore k = 60\) | M1 A1 | (10) |
**(a)** $a = \frac{dv}{dt} = 3\mathbf{i} - 2\mathbf{j}$ and when $t = 2$, $a = 3\mathbf{i} - 4\mathbf{j}$ | M1 A1 |
mag. of a = $\sqrt{(3)^2 + (-4)^2} = 5$ ms$^{-2}$ | M1 A1 |
**(b)** $s = \int v \, dt = \frac{3}{2}t^2 - \frac{1}{4}t^4\mathbf{j} + A\mathbf{i} + B\mathbf{j}$ | M1 A1 |
when $t = 0$, $s = 6\mathbf{i} + 12\mathbf{j}$ so $A = 6, B = 12$ | M1 |
$s = (\frac{3}{2}t^2 + 6)\mathbf{i} + (12 - \frac{1}{4}t^4)\mathbf{j}$ | A1 |
disp. when $t = 6$ is $60\mathbf{i} - 60\mathbf{j} = 60(\mathbf{i} - \mathbf{j}) \therefore k = 60$ | M1 A1 | (10)4. The velocity $\mathbf { v } \mathrm { ms } ^ { - 1 }$ of a particle $P$ at time $t$ seconds is given by $\mathbf { v } = 3 t \mathbf { i } - t ^ { 2 } \mathbf { j }$.
\begin{enumerate}[label=(\alph*)]
\item Find the magnitude of the acceleration of $P$ when $t = 2$.
When $t = 0$, the displacement of $P$ from a fixed origin $O$ is $( 6 \mathbf { i } + 12 \mathbf { j } ) \mathrm { m } \mathrm { s } ^ { - 1 }$, where $\mathbf { i }$ and $\mathbf { j }$ are perpendicular horizontal unit vectors.
\item Show that the displacement of $P$ from $O$ when $t = 6$ is given by $k ( \mathbf { i } - \mathbf { j } ) \mathrm { m }$, where $k$ is an integer which you should find.\\
(6 marks)
\end{enumerate}
\hfill \mbox{\textit{Edexcel M2 Q4 [10]}}