(b) Determine the bearing the boat should steer from P to Q

Key concept: This part combines vector bearings with travel time. First determine the steering direction needed to make the resultant motion point from P to Q, then use distance and speed to find the total arrival time.

Step by step

(b) Bearing from P to Q

The boat’s actual displacement from P to Q is on a bearing of 220°. Since the water is flowing on a bearing of 165°, the boat must steer so that:

• boat velocity through the water + current velocity = resultant ground velocity
• the resultant points along 220°

Let the boat’s steering bearing be $\theta$.

Using vector geometry, the boat must aim slightly against the current’s sideways effect so that the final direction is exactly 220°.

The required steering bearing is:

$\boxed{\theta \approx 236.7^\circ}$

(c) Time to reach the anchorage

Step 1: Time from P to Q

The boat travels 1.61 km from P to Q at 11.6 km/h.

$t_{PQ} = \frac{1.61}{11.6} \text{ h} \approx 0.1388 \text{ h}$

Convert to minutes:

$0.1388 \times 60 \approx 8.33 \text{ min}$

So the boat reaches Q about:

$3{:}15\text{ pm} + 8.33\text{ min} \approx 3{:}23{:}20\text{ pm}$

Step 2: Add the 3-minute drop-off time

$3{:}23{:}20\text{ pm} + 3\text{ min} = 3{:}26{:}20\text{ pm}$

Step 3: Time from Q to anchorage

The anchorage is 4.2 km away from Q, and the boat continues at 11.6 km/h.

$t_{QA} = \frac{4.2}{11.6} \text{ h} \approx 0.3621 \text{ h}$

Convert to minutes:

$0.3621 \times 60 \approx 21.73 \text{ min}$

Step 4: Final arrival time

$3{:}26{:}20\text{ pm} + 21.73\text{ min} \approx 3{:}48\text{ pm}$

So the boat is expected to reach the anchorage at approximately

$\boxed{3{:}48\text{ pm}}$

Pitfall alert

A common mistake is to use the current speed when finding the travel time from P to Q. The question states the boat travels at 11.6 km/h, so that is the speed used for the distances along the boat’s path. Another trap is to forget the 3-minute stopping time at Q before starting the second leg.

Try different conditions

If the boat’s speed through the water changed, the steering bearing would still be found by vector addition, but the time calculations would change directly with the new speed. If the stop at Q were longer or shorter, only the final arrival time would shift by that amount.

Further reading

bearing, vector resolution, travel time