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.

     (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]

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.