The distance of a ship from its harbour is modelled by the function $d(t)=-3t^{3}+3t^{2}+18t$ where t is the time elapsed in hours since departure from the harbour.

Question

a) Factor the time function.
b) When does the ship return to the harbour?
c) There is another zero of $d(t)$ . What is it, and why is it not relevant to the problem?
d) Draw a sketch of the function where $0\leqslant t\leqslant 3$
e) Estimate the time that the ship begins its return trip back to the harbour.

Answer 1

a) To factor the function $d(t)=-3t^{3}+3t^{2}+18t$, we can first factor out a common factor of $3t$ to get $d(t) = 3t(-t^2+t+6)$. Then, we can further factor the quadratic expression inside the parentheses by using the quadratic formula:

$t = \frac{-b \pm \sqrt{b^2-4ac}}{2a}$

In this case, $a=-1$, $b=1$, and $c=6$. Plugging these values into the formula, we get:

$t = \frac{-1 \pm \sqrt{1^2-4(-1)(6)}}{2(-1)}$

$t = \frac{-1 \pm \sqrt{1+24}}{-2}$

$t = \frac{-1 \pm \sqrt{25}}{-2}$

$t = \frac{-1 \pm 5}{-2}$

This gives us the factored form: $d(t) = 3t(t+1)(t-6)$

b) The ship returns to the harbour when $d(t) = 0$. So, setting $d(t)=0$, we get:

$3t(t+1)(t-6)=0$

This gives us $t = 0, -1, 6$. Since time cannot be negative, the ship returns to the harbour at $t=6$ hours.

c) The other zero of $d(t)$ is $t=-1$. However, this value is not relevant to the problem because time cannot be negative.

d)

\[
\begin{array}{|c|c|}
\hline
t & d(t)\\
\hline
0 & 0\\
1 & 18\\
2 & 60\\
3 & 114\\
\hline
\end{array}
\]

The graph of $d(t)$ is a cubic function that opens downward. We can see from the table that the ship is furthest from the harbour at $t=3$.

e) To estimate the time that the ship begins its return trip back to the harbour, we can look at the graph in part (d) and see that the ship starts heading back towards the harbour before reaching the furthest point. So, the ship likely begins its return journey around $t=2$ hours.