| Exam Board | Edexcel |
|---|---|
| Module | M1 (Mechanics 1) |
| Year | 2010 |
| Session | January |
| Marks | 14 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Vectors Introduction & 2D |
| Type | Distance between two moving objects |
| Difficulty | Moderate -0.3 This is a standard M1 kinematics question using vectors with straightforward steps: finding velocity from two positions, calculating speed and bearing, verifying a position equation, and solving a quadratic distance equation. All techniques are routine for M1 level with no novel problem-solving required, making it slightly easier than average. |
| Spec | 1.10d Vector operations: addition and scalar multiplication1.10e Position vectors: and displacement3.02a Kinematics language: position, displacement, velocity, acceleration |
| Answer | Marks | Guidance |
|---|---|---|
| speed is \(\sqrt{(3^2 + 4^2)} = 5\) (km h\(^{-1}\)) | M1 A1 M1 A1 | (4) |
| Answer | Marks | Guidance |
|---|---|---|
| bearing is 37, 36.9, 36.87, ... | M1 A1 | (2) |
| Answer | Marks | Guidance |
|---|---|---|
| \(= (3t + 9)\mathbf{i} + (4t - 6)\mathbf{j}\) ✱ | M1 A1 A1 | (2) |
| Answer | Marks | Guidance |
|---|---|---|
| \(T = 1, 5\) | M1 A1 M1 DM1 A1 A1 | (6) [14] |
**Part (a):**
$\mathbf{v} = \frac{2\mathbf{i} + 10\mathbf{j} - (9\mathbf{i} - 6\mathbf{j})}{4} = 3\mathbf{i} + 4\mathbf{j}$
speed is $\sqrt{(3^2 + 4^2)} = 5$ (km h$^{-1}$) | M1 A1 M1 A1 | (4)
**Part (b):**
$\tan\theta = \frac{3}{4}$ ($\Rightarrow \theta \approx 36.9°$)
bearing is 37, 36.9, 36.87, ... | M1 A1 | (2)
**Part (c):**
$\mathbf{s} = 9\mathbf{i} - 6\mathbf{j} + t(3\mathbf{i} + 4\mathbf{j})$
$= (3t + 9)\mathbf{i} + (4t - 6)\mathbf{j}$ ✱ | M1 A1 A1 | (2) | cso
**Part (d):**
Position vector of S relative to L is $(3T + 9)\mathbf{i} + (4T - 6)\mathbf{j} - (-18\mathbf{i} + 6\mathbf{j}) = (3T - 9)\mathbf{i} + (4T - 12)\mathbf{j}$
$(3T - 9)^2 + (4T - 12)^2 = 100$
$25T^2 - 150T + 125 = 0$ or equivalent
($T^2 - 6T + 5 = 0$)
$T = 1, 5$ | M1 A1 M1 DM1 A1 A1 | (6) [14][In this question, $\mathbf{i}$ and $\mathbf{j}$ are horizontal unit vectors due east and due north respectively and position vectors are given with respect to a fixed origin.]
A ship $S$ is moving along a straight line with constant velocity. At time $t$ hours the position vector of $S$ is $\mathbf{s}$ km. When $t = 0$, $\mathbf{s} = 9\mathbf{i} - 6\mathbf{j}$. When $t = 4$, $\mathbf{s} = 21\mathbf{i} + 10\mathbf{j}$. Find
\begin{enumerate}[label=(\alph*)]
\item the speed of $S$, [4]
\item the direction in which $S$ is moving, giving your answer as a bearing. [2]
\item Show that $\mathbf{s} = (3t + 9)\mathbf{i} + (4t - 6)\mathbf{j}$. [2]
\end{enumerate}
A lighthouse $L$ is located at the point with position vector $(18\mathbf{i} + 6\mathbf{j})$ km. When $t = T$, the ship $S$ is 10 km from $L$.
\begin{enumerate}[label=(\alph*)]
\setcounter{enumi}{3}
\item Find the possible values of $T$. [6]
\end{enumerate}
\hfill \mbox{\textit{Edexcel M1 2010 Q7 [14]}}