Relative velocity with a constant crosswind or current: adding velocity vectors, resultant speed and direction, the heading needed to hold a course, and bearings (new in the 2024 Extension 1 syllabus)

A wind from the north blows towards the south. Ground velocity (east, north) $= (250, -50)$.

Ground speed $= \sqrt{250^2 + 50^2} = \sqrt{65\,000} \approx 255.0$ km/h.

The track is south of east by $\tan^{-1}\frac{50}{250} \approx 11.3^\circ$, so the true bearing is $90^\circ + 11.3^\circ = 101.3^\circ$.

Markers expect the correct interpretation of "from the north", a vector (or labelled triangle) sum, the speed, and a bearing measured clockwise from north.

The upstream component must cancel the current: $5\sin\theta = 2$, so $\sin\theta = 0.4$ and $\theta \approx 23.6^\circ$ upstream of straight across.

Speed across $= \sqrt{5^2 - 2^2} = \sqrt{21} \approx 4.58$ km/h.

Markers award one mark for the cancellation condition, one for the angle with its reference direction, and one for the speed.

The two velocities are perpendicular, so the resultant speed is

Angle downstream of straight across. $\tan\theta = \frac{3}{4}$, so $\theta \approx 36.9^\circ$.

Marker's note: one mark for $5$ m/s, one for the angle with its reference direction stated ("downstream of straight across").

Add the vectors. Air velocity $200\mathbf{j}$, wind $30\mathbf{i}$, so the ground velocity is $30\mathbf{i} + 200\mathbf{j}$.

Magnitude. $\sqrt{30^2 + 200^2} = \sqrt{40\,900} \approx 202.2$ km/h.

Marker's note: one mark for the vector sum, one for the speed.

Time

Only the across component carries the boat over: $t = \frac{80}{2} = 40$ s.

Drift

In that time the current carries it $1.5 \times 40 = 60$ m downstream.

Speed

$\sqrt{2^2 + 1.5^2} = 2.5$ m/s.

Marker's note: one mark for each result. Dividing $80$ by the resultant speed $2.5$ is the standard error: the path is longer than $80$ m.

Aim upstream. Let the heading make angle $\theta$ upstream of straight across. The upstream component of the boat's velocity must cancel the current:

Speed across. $\sqrt{2^2 - 1.5^2} = \sqrt{1.75} \approx 1.323$ m/s.

Time. $\frac{80}{\sqrt{1.75}} \approx 60.5$ s.

Marker's note: one mark for the heading (with "upstream of straight across" stated), one for the across speed, one for the time.

Wind from the west blows towards the east, so the ground velocity is $60\mathbf{i} + 400\mathbf{j}$ (east, north).

Ground speed. $\sqrt{60^2 + 400^2} = \sqrt{163\,600} \approx 404.5$ km/h.

Direction. The track is east of north by $\tan^{-1}\frac{60}{400} \approx 8.5^\circ$, so the true bearing is about $008.5^\circ$ (N $8.5^\circ$ E).

Marker's note: one mark for interpreting "from the west", one for the speed, one for the bearing.

Heading

Steer west of north by $\alpha$ so the westward component of the air velocity cancels the wind: $300\sin\alpha = 50$, so $\sin\alpha = \frac{1}{6}$ and $\alpha \approx 9.59^\circ$. The heading is $360^\circ - 9.59^\circ \approx 350.4^\circ$ true.

Ground speed

The northward component is $300\cos\alpha = \sqrt{300^2 - 50^2} = \sqrt{87\,500} \approx 295.8$ km/h.

Time

$\frac{500}{295.8} \approx 1.690$ hours, which is $1$ hour $41$ minutes.

Marker's note: one mark for the cancellation condition, one for the bearing, one for the ground speed, one for the time.

Components (east, north). A velocity of speed $v$ towards bearing $\beta$ has components $(v\sin\beta, v\cos\beta)$.

Air velocity: $(250, 0)$. Wind: $(40\sin 45^\circ, 40\cos 45^\circ) = (20\sqrt{2}, 20\sqrt{2}) \approx (28.28, 28.28)$.

Ground velocity

$(278.28, 28.28)$.

Speed

$\sqrt{278.28^2 + 28.28^2} \approx 279.7$ km/h.

Bearing

The track is north of east by $\tan^{-1}\frac{28.28}{278.28} \approx 5.8^\circ$, so the bearing is $90^\circ - 5.8^\circ \approx 084^\circ$.

Marker's note: one mark for resolving the wind, one for the sum, one for the speed, one for the bearing. Using the cosine rule on the vector triangle (angle $135^\circ$ between the two arrows placed tip to tail) gives the same speed.

Set up components. Take $x$ downstream and $y$ across. Heading $\theta$ upstream of straight across, the ground velocity is

Direction condition. To land $20$ m downstream after $60$ m across, $\frac{1 - 1.5\sin\theta}{1.5\cos\theta} = \frac{20}{60} = \frac{1}{3}$, so

Auxiliary angle. $3\sin\theta + \cos\theta = \sqrt{10}\sin(\theta + \alpha)$ with $\tan\alpha = \frac{1}{3}$, $\alpha \approx 18.43^\circ$. Then $\sin(\theta + \alpha) = \frac{2}{\sqrt{10}}$, so $\theta + \alpha \approx 39.23^\circ$ (the other solution gives a heading with a negative across component) and $\theta \approx 20.8^\circ$.

Time. $\frac{60}{1.5\cos 20.8^\circ} \approx 42.8$ s.

Marker's note: one mark for the component form, one for the direction equation, two for solving it (auxiliary angle and rejecting the extra root), one for the time.