Port A is defined to be the origin of a set of coordinate axes and port B is located at the point (70, 30), where distances are measured in kilometres. A ship S₁ sails from port A at 10:00 in a straight line such that its position \(t\) hours after 10:00 is given by \(r = t\left( {\begin{array}{*{20}{c}}
  {10} \\
  {20}
\end{array}} \right)\).
A speedboat S₂ is capable of three times the speed of S₁ and is to meet S₁ by travelling the shortest possible distance. What is the latest time that S₂ can leave port B?

METHOD 1

equation of journey of ship S₁

\({r_1} = t\left( {\begin{array}{*{20}{c}}
  {10} \\
  {20}
\end{array}} \right)\)

equation of journey of speedboat S₂ ,setting off \(k\) minutes later

\({r_2} = \left( {\begin{array}{*{20}{c}}
  {70} \\
  {30}
\end{array}} \right) + \left( {t - k} \right)\left( {\begin{array}{*{20}{c}}
  { - 60} \\
  {30}
\end{array}} \right)\)     M1A1A1

Note: Award M1 for perpendicular direction, A1 for speed, A1 for change in parameter (e.g. by using \(t - k\) or \(T\), \(k\) being the time difference between the departure of the ships).

solve \(t\left( {\begin{array}{*{20}{c}}
  {10} \\
  {20}
\end{array}} \right) = \left( {\begin{array}{*{20}{c}}
  {70} \\
  {30}
\end{array}} \right) + \left( {t - k} \right)\left( {\begin{array}{*{20}{c}}
  { - 60} \\
  {30}
\end{array}} \right)\)     (M1)

Note: M mark is for equating their two expressions.

\(10t = 70 - 60t + 60k\)

\(20t = 30 + 30t - 30k\)     M1

Note: M mark is for obtaining two equations involving two different parameters.

\(7t - 6k = 7\)

\( - t + 3k = 3\)

\(k = \frac{{28}}{{15}}\)     A1

latest time is 11:52     A1

[7 marks]

METHOD 2

\({\text{SB}} = 22\sqrt 5 \)     M1A1
(by perpendicular distance)

\({\text{SA}} = 26\sqrt 5 \)     M1A1
(by Pythagoras or coordinates)

\(t = \frac{{26\sqrt 5 }}{{10\sqrt 5 }}\)     A1

\(t - k = \frac{{22\sqrt 5 }}{{30\sqrt 5 }}\)     A1

\(k = \frac{{28}}{{15}}\) leading to latest time 11:52     A1

[7 marks]

Few candidates managed to make progress on this question. Many candidates did not attempt the problem and many that did make an attempt failed to draw a diagram that would have allowed them to make further progress. There were a variety of possible solution techniques but candidates seemed unable to interpret the equation of a straight line written in vector form or find a perpendicular direction. This meant that it was very difficult for meaningful progress to be made towards a solution.