| Exam Board | Edexcel |
|---|---|
| Module | AS Paper 1 (AS Paper 1) |
| Year | 2020 |
| Session | June |
| Marks | 6 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Vectors Introduction & 2D |
| Type | Bearing and speed from velocity vector |
| Difficulty | Moderate -0.8 This is a straightforward application of basic vector arithmetic and trigonometry. Students find the displacement vector by subtraction, calculate bearing using arctan (with careful quadrant consideration), and find speed by dividing distance by time. All steps are routine with no problem-solving insight required, making it easier than average but not trivial due to the bearing calculation requiring some care. |
| Spec | 1.10d Vector operations: addition and scalar multiplication1.10e Position vectors: and displacement1.10f Distance between points: using position vectors1.10h Vectors in kinematics: uniform acceleration in vector form |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| Attempts to find an allowable angle e.g. \(\tan\theta = \frac{7}{3}\) | M1 | Either "66.8°", "23.2°" or ("49.8°" and "63.4°"). Must attempt to subtract coordinates. If part (b) attempted first, allow \(\sin\theta = \pm\frac{7}{"\sqrt{58}"}\), \(\cos\theta = \pm\frac{7}{"\sqrt{58}"}\) etc |
| Full attempt to find bearing e.g. \(180° + {"}67°{"}\) | dM1 | \(180° + \arctan\frac{7}{3}\), \(270° - \arctan\frac{3}{7}\), \(360° - {"}49.8°{"} - {"}63.4°{"}\). Dependent on previous M1 |
| Bearing \(= \text{awrt } 246.8°\) | A1 | Allow S \(66.8°\) W |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| Distance \(= \sqrt{(4--3)^2+(-2+5)^2} = \sqrt{58}\) | M1 | Allow \(d^2 = (4--3)^2+(-2+5)^2\). May be seen on diagram using resultant vector from (a) |
| Speed \(= \frac{\sqrt{58}}{2.75}\) | dM1 | Must have attempted distance using coordinates then divide by 2.75. Alternatively find speed in km min\(^{-1}\) then multiply by 60 |
| \(= \text{awrt } 2.77 \text{ km h}^{-1}\) | A1 |
## Question 2:
### Part (a):
| Answer/Working | Mark | Guidance |
|---|---|---|
| Attempts to find an allowable angle e.g. $\tan\theta = \frac{7}{3}$ | M1 | Either "66.8°", "23.2°" or ("49.8°" and "63.4°"). Must attempt to subtract coordinates. If part (b) attempted first, allow $\sin\theta = \pm\frac{7}{"\sqrt{58}"}$, $\cos\theta = \pm\frac{7}{"\sqrt{58}"}$ etc |
| Full attempt to find bearing e.g. $180° + {"}67°{"}$ | dM1 | $180° + \arctan\frac{7}{3}$, $270° - \arctan\frac{3}{7}$, $360° - {"}49.8°{"} - {"}63.4°{"}$. Dependent on previous M1 |
| Bearing $= \text{awrt } 246.8°$ | A1 | Allow S $66.8°$ W |
### Part (b):
| Answer/Working | Mark | Guidance |
|---|---|---|
| Distance $= \sqrt{(4--3)^2+(-2+5)^2} = \sqrt{58}$ | M1 | Allow $d^2 = (4--3)^2+(-2+5)^2$. May be seen on diagram using resultant vector from (a) |
| Speed $= \frac{\sqrt{58}}{2.75}$ | dM1 | Must have attempted distance using coordinates then divide by 2.75. Alternatively find speed in km min$^{-1}$ then multiply by 60 |
| $= \text{awrt } 2.77 \text{ km h}^{-1}$ | A1 | |
---\begin{enumerate}
\item \hspace{0pt} [In this question the unit vectors $\mathbf { i }$ and $\mathbf { j }$ are due east and due north respectively.]
\end{enumerate}
A coastguard station $O$ monitors the movements of a small boat.\\
At 10:00 the boat is at the point $( 4 \mathbf { i } - 2 \mathbf { j } ) \mathrm { km }$ relative to $O$.\\
At 12:45 the boat is at the point $( - 3 \mathbf { i } - 5 \mathbf { j } ) \mathrm { km }$ relative to $O$.\\
The motion of the boat is modelled as that of a particle moving in a straight line at constant speed.\\
(a) Calculate the bearing on which the boat is moving, giving your answer in degrees to one decimal place.\\
(b) Calculate the speed of the boat, giving your answer in $\mathrm { kmh } ^ { - 1 }$
\hfill \mbox{\textit{Edexcel AS Paper 1 2020 Q2 [6]}}