| 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 -0.3 This is a straightforward M2 kinematics question requiring substitution to find k, then differentiation to find velocity and acceleration vectors. The concepts are standard (position→velocity→acceleration) with routine vector arithmetic and no problem-solving insight needed. Slightly easier than average due to the direct application of techniques. |
| Spec | 1.10h Vectors in kinematics: uniform acceleration in vector form3.02a Kinematics language: position, displacement, velocity, acceleration |
| Answer | Marks | Guidance |
|---|---|---|
| (a) When \(t = 4\), \(r = 12i + (16k - 5)j\) | M1 A1 A1 | |
| \(16k - 5 = 1\) | ||
| \(k = 0.375\) | M1 A1 A1 | |
| (b) \(v = 2i + 0.75j\) | M1 A1 A1 A1 | |
| \(a = 0.75j\) | ||
| \( | a | = 0.75\) m/s\(^{-2}\), due North |
**(a)** When $t = 4$, $r = 12i + (16k - 5)j$ | M1 A1 A1 |
$16k - 5 = 1$ | |
$k = 0.375$ | M1 A1 A1 |
**(b)** $v = 2i + 0.75j$ | M1 A1 A1 A1 |
$a = 0.75j$ | |
$|a| = 0.75$ m/s$^{-2}$, due North | M1 A1 A1 A1 | **Total: 7 marks**Relative to a fixed origin $O$, the points $X$ and $Y$ have position vectors $(4\mathbf{i} - 5\mathbf{j})$ m and $(12\mathbf{i} + \mathbf{j})$ m respectively, where $\mathbf{i}$ and $\mathbf{j}$ are perpendicular unit vectors in the directions due east and due north respectively. A particle $P$ starts from $X$, and $t$ seconds later its position vector relative to $O$ is $(2t + 4)\mathbf{i} + (kt^2 - 5)\mathbf{j}$.
\begin{enumerate}[label=(\alph*)]
\item Find the value of $k$ if $P$ takes $4$ seconds to reach $Y$. [3 marks]
\item Show that $P$ has constant acceleration and find the magnitude and direction of this acceleration. [4 marks]
\end{enumerate}
\hfill \mbox{\textit{Edexcel M2 Q4 [7]}}