M4 June 2018 Q5
5. A horizontal square field, \(PQRS\), has sides of length 75 m. Ali is at corner \(P\) of the field and Beth is at corner \(Q\) of the field. Ali starts to walk in a straight line along the diagonal of the field from \(P\) to \(R\) at a constant speed of 1.5 m s\(^{-1}\). Beth sees Ali start to walk, waits 10 seconds, and then walks from \(Q\) to intercept Ali. Beth walks in a straight line at a constant speed of 2 m s\(^{-1}\).
Find
(i) the time from the instant Beth leaves \(Q\) until the instant that she intercepts Ali,
(ii) the direction Beth should take. (11)
| Scheme | Marks |
|---|---|
| Using distances | |
| Distance travelled by Ali: \(1.5(t + 10)\) | M1A1 |
| Distance travelled by Beth: \(2t\) and correct triangle seen or implied | B1 |
| Cosine rule: \((2t)^2 = 75^2 + 1.5^2(t + 10)^2 - 2\times 75\times 1.5(t + 10)\cos 45\) | M1A1 |
| \(1.75t^2 + 114.1t - 4259 = 0\) | M1 |
| \(t = \dfrac{-114.1 \pm \sqrt{114.1^2 + 4\times 1.75\times 4259}}{3.5} = 26.5\) | A1 |
| Sine rule: \(\dfrac{\sin\alpha}{1.5(t + 10)} = \dfrac{\sin 45}{2t}\) | M1A1 |
| \(\dfrac{\sin\alpha}{1.5\times 36.5} = \dfrac{\sin 45}{2\times 26.5}\ \Rightarrow \alpha = 46.9^\circ\) to side \(PQ\) or equivalent | M1A1 |
| (11) | |
| (11 marks) |
5 Alt
| Position vector of \(A\): \(\begin{pmatrix}\frac{1.5}{\sqrt{2}}(t + 10)\\[4pt]\frac{1.5}{\sqrt{2}}(t + 10)\end{pmatrix}\) or with \(t\) | B1 |
| Position vector of \(B\): \(\begin{pmatrix}2t\sin\alpha\\75 - 2t\cos\alpha\end{pmatrix}\) value for time consistent | M1A1 |
| Equate components: | M1A1 |
| Form equation in t: \(4t^2 = \dfrac{9}{4}\times\dfrac{1}{2}(t + 10)^2 + \left(75 - \dfrac{3}{2\sqrt{2}}(t + 10)\right)^2\) | M1A1 |
| Simplify and solve: \(14t^2 + 912.8t - 34072 = 0\) | M1 |
| \(t = 26.5\) | A1 |
| Substitute \(t\) and solve for \(\alpha\) | M1 |
| \(\Rightarrow \alpha = 46.9^\circ\) to side \(PQ\) | A1 |
| (11) |
(Corrected from the printed mark scheme: the components of the position vector of \(A\) are printed as \(\dfrac{1.5}{2}(t + 10)\), and the bracket in the equation in \(t\) as \((t - 10)\).)
5 Alt – Using triangle of velocities
| Using distances: \(\tan\theta = \dfrac{15/\sqrt{2}}{75 - 15/\sqrt{2}}\qquad \theta = 9.35^\circ\) | M1A1 |
| \(\alpha = 45^\circ + \theta = 54.35^\circ\) | |
| Distance to travel at relative velocity: \(\sqrt{\left(15/\sqrt{2}\right)^2 + \left(75 - 15/\sqrt{2}\right)^2} = \sqrt{10.61^2 + 64.39^2} = 65.3\) (m) | B1 |
| Using relative velocities: \(\dfrac{\sin\alpha}{2} = \dfrac{\sin\beta}{1.5}\) their \(\alpha, \beta\) | M1A1 |
| \(\beta = 37.5^\circ\) | |
| \(\Rightarrow\) Beth should travel at \(\theta + \beta = 46.9^\circ\) to side \(PQ\) or equivalent | M1A1 |
| Relative velocity: \(\dfrac{v}{\sin(180 - \alpha - \beta)} = \dfrac{2}{\sin\alpha}\) | M1A1 |
| \(v = 2.46\) (m s\(^{-1}\)) | |
| Time to intercept \(= \dfrac{65.3}{2.46} = 26.5\) (s) | M1A1 |
| (11) |