| Exam Board | Edexcel |
|---|---|
| Module | M2 (Mechanics 2) |
| Marks | 7 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Variable acceleration (1D) |
| Type | Vector motion with components |
| Difficulty | Moderate -0.8 This is a straightforward M2 kinematics question requiring only routine differentiation of vectors and basic magnitude calculations. Part (a) is direct differentiation, part (b) requires showing the acceleration vector is constant (one more differentiation), and part (c) involves a simple magnitude calculation and solving a linear equation. No problem-solving insight needed, just mechanical application of standard techniques. |
| Spec | 1.07b Gradient as rate of change: dy/dx notation1.10h Vectors in kinematics: uniform acceleration in vector form3.02g Two-dimensional variable acceleration |
| Answer | Marks | Guidance |
|---|---|---|
| (a) \(\mathbf{v} = e^t \mathbf{i} - 2\mathbf{j}\) | M1 A1 | |
| (b) \(\mathbf{a} = e^t \mathbf{i}\), so always in i-direction | M1 A1 | |
| (c) When \( | \mathbf{a} | = 12\), \(t = \ln 12 = 2.48 \text{ s}\) |
**(a)** $\mathbf{v} = e^t \mathbf{i} - 2\mathbf{j}$ | M1 A1 |
**(b)** $\mathbf{a} = e^t \mathbf{i}$, so always in i-direction | M1 A1 |
**(c)** When $|\mathbf{a}| = 12$, $t = \ln 12 = 2.48 \text{ s}$ | M1 A1 A1 | 7 marksA particle $P$ moves in a plane such that its position vector $\mathbf{r}$ metres at time $t$ seconds, relative to a fixed origin $O$, is $\mathbf{r} = t^2\mathbf{i} - 2t\mathbf{j}$.
\begin{enumerate}[label=(\alph*)]
\item Find the velocity vector of $P$ at time $t$ seconds. [2 marks]
\item Show that the direction of the acceleration of $P$ is constant. [2 marks]
\item Find the value of $t$ when the acceleration of $P$ has magnitude 12 ms$^{-2}$. [3 marks]
\end{enumerate}
\hfill \mbox{\textit{Edexcel M2 Q3 [7]}}