A ferry company runs a service between two ports that are 84 km apart. On Monday a ferry made the crossing at an average speed of \(v\) km/h. On Tuesday the...

Two journeys over the same distance at different speeds, with a stated difference in time, always lead to a quadratic. The relation used throughout is \(\text{time} = \dfrac{\text{distance}}{\text{speed}}\).

(a) Time on Monday. [1]

\[t_{\text{Mon}} = \frac{84}{v}\ \text{hours}\]

(b) Show that \(v^2 + 7v - 588 = 0\). [4]
Tuesday's speed is \((v + 7)\) km/h, so Tuesday's time is \(\dfrac{84}{v + 7}\) hours. Tuesday was the faster crossing, so it was the shorter one, and the difference is 1 hour:

\[\frac{84}{v} - \frac{84}{v + 7} = 1\]

Multiply every term by \(v(v + 7)\):

\[84(v + 7) - 84v = v(v + 7)\] \[84v + 588 - 84v = v^2 + 7v\] \[588 = v^2 + 7v \quad\Rightarrow\quad v^2 + 7v - 588 = 0\]

Marks: one for Tuesday's time, one for the equation with the subtraction the right way round, one for clearing fractions, one for the final form. Subtracting in the wrong order gives \(-1\) on the right and leads to \(v^2 + 7v + 588 = 0\), which has no real roots at all.

(c) Solve for \(v\). [3]
Using the formula with \(a = 1\), \(b = 7\), \(c = -588\):

\[b^2 - 4ac = 49 + 2352 = 2401, \qquad \sqrt{2401} = 49\] \[v = \frac{-7 \pm 49}{2}\] \[v = \frac{42}{2} = 21 \qquad\text{or}\qquad v = \frac{-56}{2} = -28\]

A speed cannot be negative, so \(v = 21\) km/h.

(d) Time on Tuesday. [2]

\[\text{speed} = 21 + 7 = 28\ \text{km/h}, \qquad t = \frac{84}{28} = 3\ \text{hours}\]

Check the story: Monday took \(\frac{84}{21} = 4\) hours, and \(4 - 3 = 1\) hour, exactly the stated difference.

(e) Average speed of the Islander. [2]
On the distance-time graph the line for the Islander runs from the origin up to 84 km at 3.5 hours. Its gradient is the average speed:

\[\text{speed} = \frac{84}{3.5} = 24\ \text{km/h}\]

(f) Does the Islander meet the rule? [3]
The rule is that a crossing must take less than 3 hours 30 minutes, that is less than 3.5 hours. The Islander takes exactly 3.5 hours, which is not less than 3.5, so it does not meet the rule.

For a crossing under 3.5 hours the speed must satisfy

\[\frac{84}{v} < 3.5 \quad\Rightarrow\quad 84 < 3.5v \quad\Rightarrow\quad v > 24\]

The least whole number speed that works is 25 km/h, which gives a crossing time of \(\frac{84}{25} = 3.36\) hours, that is 3 hours 21.6 minutes.

Examination point: "less than" excludes the boundary value. An exact 3.5 hour crossing fails a "less than 3.5 hours" rule, and \(v > 24\) means 24 km/h itself is not allowed, so the least whole number is 25 and not 24.