| Date | November 2011 | Marks available | 8 | Reference code | 11N.1.hl.TZ0.11 |
| Level | HL only | Paper | 1 | Time zone | TZ0 |
| Command term | Determine | Question number | 11 | Adapted from | N/A |
Question
At 12:00 a boat is 20 km due south of a freighter. The boat is travelling due east at \(20{\text{ km}}\,{{\text{h}}^{ - 1}}\), and the freighter is travelling due south at \(40{\text{ km}}\,{{\text{h}}^{ - 1}}\).
Determine the time at which the two ships are closest to one another, and justify your answer.
If the visibility at sea is 9 km, determine whether or not the captains of the two ships can ever see each other’s ship.
Markscheme
(M1)
\({s^2} = {(20t)^2} + {(20 - 40t)^2}\) M1
\({s^2} = 2000{t^2} - 1600t + 400\) A1
to minimize s it is enough to minimize \({s^2}\)
\(f'(t) = 4000t - 1600\) A1
setting \(f'(t)\) equal to 0 M1
\(4000t - 1600 = 0 \Rightarrow t = \frac{2}{5}\) or 24 minutes A1
\(f''(t) = 4000 > 0\) M1
\( \Rightarrow \) at \(t = \frac{2}{5},{\text{ }}f(t)\) is minimized
hence, the ships are closest at 12:24 A1
Note: accept solution based on s.
[8 marks]
\(f\left( {\frac{2}{5}} \right) = \sqrt {80} \) M1A1
since \(\sqrt {80} < 9\), the captains can see one another R1
[3 marks]
Examiners report
This was, disappointingly, a poorly answered question. Some tried to talk their way through the question without introducing the time variable. Even those who did use the distance as a function of time often did not check for a minimum.
This was, disappointingly, a poorly answered question. Some tried to talk their way through the question without introducing the time variable. Even those who did use the distance as a function of time often did not check for a minimum.
