M4 June 2016 Q3
3. Two straight horizontal roads cross at right angles at the point \(X\). A girl is running, with constant speed 5 m s\(^{-1}\), due north towards \(X\) on one road. A car is travelling, with constant speed 20 m s\(^{-1}\), due west towards \(X\) on the other road.
At noon the girl is 150 m south of \(X\) and the car is 800 m east of \(X\).
| Scheme | Marks |
|---|---|
| \({}_{c}\mathbf{v}_{g} = \mathbf{v}_{c} - \mathbf{v}_{g}\) | B1 |
| \(\tan\theta = \dfrac{20}{5}\qquad \theta = 75.96^\circ\) or \(\tan\theta = \dfrac{5}{20},\quad \theta = 14.04^\circ\) | M1 |
| A1 | |
| Direction is \(256^\circ\) | A1 |
| Mag \(= \sqrt{20^2 + 5^2}\) | M1 |
| \(= \sqrt{425}\qquad (= 20.61\ldots)\) (m s\(^{-1}\)) | A1 |
| (6) |
Notes
B1 Correct vector triangle seen or implied e.g. sight of \({}_{c}\mathbf{v}_{g} = \begin{pmatrix}-20\\-5\end{pmatrix}\) or correct final bearing
M1 Use trig. to find a relevant angle
A1 Angle correct
A1 \(5\sqrt{17}\) Accept 21
(Corrected from the printed mark scheme: the magnitude is printed as \(\sqrt{20^2 + 5^5}\).)
| Scheme | Marks |
|---|---|
| Dist apart at noon \(= \sqrt{150^2 + 800^2}\ \left(= \sqrt{662500} = 813.94\ldots\right)\) | M1 |
| \(\tan\alpha = \dfrac{150}{800}\) | M1 |
| \(\alpha = \tan^{-1}\left(\dfrac{150}{800}\right),\quad (\alpha = 10.619)\) | A1 |
| \(\beta = 14.04 - 10.619 = 3.420\ldots\) | M1 (A1) |
| \(\sin\beta = \dfrac{d}{\sqrt{662500}}\) | M1 |
| A1 | |
| \(d = \sqrt{662500}\sin 3.42\ = 48.5\ldots\) m | A1 |
| (7) | |
| (13 marks) |
Notes
M1 Use trig to find \(\alpha\)
A1 Correct equation in \(\alpha\)
M1 (A1) Correct strategy for \(\beta\). Their \(\theta\) – their \(\alpha\)
M1 Use trig. to find \(d\)
A1 Correct unsimplified expression
A1 or exact answer \(\dfrac{200}{\sqrt{17}}\). Accept 49 or better
3b alt
| Relative position \(\begin{pmatrix}800-20t\\150-5t\end{pmatrix}\) | M1 |
| Distance \(d^2 = (800-20t)^2 + (150-5t)^2\) | M1 |
| \((= 425t^2 - 33500t + 662500)\) | A1 |
| \(-40(800-20t) - 10(150-5t)\) | M1 |
| \((850t - 33500 = 0)\) | M1 (A1) |
| \(t = \dfrac{670}{17},\ 39.4\) (s) | A1 |
| \(\Rightarrow d = 48.5\) (m) | A1 |
M1 By subtraction
M1 Correct use of Pythagoras' theorem
A1 Correct unsimplified expression for \(d\) or \(d^2\)
M1 Differentiate
M1 (A1) Equate to zero and solve for \(t\).
3b alt
| Relative position \(\begin{pmatrix}800-20t\\150-5t\end{pmatrix}\) | M1 |
| Distance \(d^2 = (800-20t)^2 + (150-5t)^2\) | M1 |
| \((= 425t^2 - 33500t + 662500)\) | A1 |
| \(\begin{pmatrix}800-20t\\150-5t\end{pmatrix}\cdot\begin{pmatrix}-20\\-5\end{pmatrix} = 0\) | M1 |
| \(-20(800-20t) - 5(150-5t) = 0\) | M1 |
| \(t = \dfrac{670}{17},\ 39.4\) (s) | A1 |
| \(\Rightarrow d = 48.5\) (m) | A1 |
M1 By subtraction
M1 Correct use of Pythagoras' theorem
A1 Correct unsimplified expression for \(d\) or \(d^2\)
M1 Use scalar product with \({}_{c}\mathbf{v}_{g} = \begin{pmatrix}-20\\-5\end{pmatrix}\)
M1 Equate scalar product to zero and solve for \(t\)