| Exam Board | Edexcel |
|---|---|
| Module | M1 (Mechanics 1) |
| Marks | 11 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Vectors Introduction & 2D |
| Type | Position vector at time t (constant velocity) |
| Difficulty | Moderate -0.8 This is a straightforward M1 mechanics question requiring basic vector arithmetic (finding displacement, position at a given time), speed calculation from distance/time, and bearing from vector components. All steps are routine applications of standard formulas with no conceptual challenges or problem-solving insight required. |
| Spec | 1.10c Magnitude and direction: of vectors1.10e Position vectors: and displacement3.02a Kinematics language: position, displacement, velocity, acceleration |
| Answer | Marks | Guidance |
|---|---|---|
| (a) displacement of plane \(= (32i + 19j) - (80i + 5j) = 48i + 14j\) in 10 mins. | M1 A1 | |
| \(\therefore\) in 30 mins, displacement \(= 3 \times (48i + 14j) = 144i + 42j\) | M1 | |
| so posn. vector at 2:30p.m. is \((-64i + 47j)\) | A1 | |
| (b) in 1 hr. displacement of plane \(= 6 \times (48i + 14j) = 288i + 84j\) | M1 A1 | |
| speed \(= \sqrt{(288)^2 + 84^2} = \sqrt{90000} = 300 \text{ kmh}^{-1}\) | M1 A1 | |
| (c) [diagram showing right triangle with vertical side 14 and horizontal side 48] | ||
| req'd angle \(= \tan^{-1}\frac{14}{48} = 16.26°\) | M1 A1 | |
| \(\therefore\) bearing \(= 16.26 + 270 = 286°\) (nearest deg) | A1 | (11) |
**(a)** displacement of plane $= (32i + 19j) - (80i + 5j) = 48i + 14j$ in 10 mins. | M1 A1 |
$\therefore$ in 30 mins, displacement $= 3 \times (48i + 14j) = 144i + 42j$ | M1 |
so posn. vector at 2:30p.m. is $(-64i + 47j)$ | A1 |
**(b)** in 1 hr. displacement of plane $= 6 \times (48i + 14j) = 288i + 84j$ | M1 A1 |
speed $= \sqrt{(288)^2 + 84^2} = \sqrt{90000} = 300 \text{ kmh}^{-1}$ | M1 A1 |
**(c)** [diagram showing right triangle with vertical side 14 and horizontal side 48] |
req'd angle $= \tan^{-1}\frac{14}{48} = 16.26°$ | M1 A1 |
$\therefore$ bearing $= 16.26 + 270 = 286°$ (nearest deg) | A1 | (11)4. The position of an aeroplane flying in a straight horizontal line at constant speed is plotted on a radar screen. At 2 p.m. the position vector of the aeroplane is $( 80 \mathbf { i } + 5 \mathbf { j } )$, where $\mathbf { i }$ and $\mathbf { j }$ are unit vectors directed east and north respectively relative to a fixed origin, $O$, on the screen. Ten minutes later the position of the aeroplane on the screen is $( 32 \mathbf { i } + 19 \mathbf { j } )$.
Each unit on the screen represents 1 km .
\begin{enumerate}[label=(\alph*)]
\item Find the position vector of the aeroplane at 2:30 p.m.
\item Find the speed of the aeroplane in $\mathrm { km } \mathrm { h } ^ { - 1 }$.
\item Find, correct to the nearest degree, the bearing on which the aeroplane is flying.
\end{enumerate}
\hfill \mbox{\textit{Edexcel M1 Q4 [11]}}