Aircraft and Crosswind Problems: Original Physics Practice Exam

This original tutorgum practice paper tests aircraft and crosswind problems, including the difference between heading and ground track, airspeed and ground speed, choosing a heading to compensate for wind, and calculating wind or airspeed from a specified ground journey. It uses the supplied scope compatible with the Physics 2017 and Physics 2025 syllabus editions. No other Physics topics are assessed. This is an original practice paper, not an official, endorsed, past, or predicted NESA paper.

01Exam instructions

• Suggested reading time: 5 minutes.
• Suggested working time: 50 minutes.
• Total marks: 40.
• Permitted equipment: Scientific calculator, pen or pencil, and ruler if desired. No notes.
• Working: Show your calculations, substitutions, units, and directions. A numerical answer without supporting working may not receive full marks.

These are suggested conditions for this practice paper, not claims about official examination rules.

Conventions and equations

Use the vector relationship

\[ \vec v_g=\vec v_a+\vec v_w \]

where:

• \(\vec v_g\) is the aircraft’s ground velocity.
• \(\vec v_a\) is the aircraft’s velocity relative to the surrounding air.
• \(\vec v_w\) is the wind velocity relative to the ground.

All velocities are measured in km/h unless stated otherwise.

Take east and north as the positive component directions. Bearings are measured clockwise from north: \(000^\circ\) is north, \(090^\circ\) is east, \(180^\circ\) is south, and \(270^\circ\) is west.

For a velocity with speed \(v\) on bearing \(\theta\), its components are

\[ v_E=v\sin\theta \]

\[ v_N=v\cos\theta \]

A wind described as coming from the west moves towards the east. Treat all stated velocities as constant during each flight.

Round final speeds to 3 significant figures and bearings or angles to the nearest degree, unless otherwise specified. Keep unrounded values during intermediate calculations.

02Questions

Question 1 (6 marks)

An aircraft flies with an airspeed of \(180\text{ km/h}\), heading due north. A steady wind blows from the west at \(40\text{ km/h}\).

(a) Distinguish between heading and ground track, and between airspeed and ground speed. (2 marks)

(b) Calculate the aircraft’s ground speed and ground-track bearing. (4 marks)

Question 2 (7 marks)

A light aircraft has an airspeed of \(200\text{ km/h}\). The wind velocity is \(50\text{ km/h}\) due east. The pilot wants the aircraft to maintain a ground track due north.

(a) Determine the required heading angle relative to north. State whether the aircraft must point east or west of north. (3 marks)

(b) Give the required heading as a three-digit bearing, to the nearest degree. (1 mark)

(c) Calculate the ground speed along the required track. (3 marks)

Question 3 (6 marks)

During a steady flight, an aircraft has a heading due east and an airspeed of \(150\text{ km/h}\). Its ground speed is measured as \(130\text{ km/h}\), on a ground track \(10.0^\circ\) south of east.

(a) Calculate the east and north components of the aircraft’s ground velocity. (2 marks)

(b) Determine the east and north components of the wind velocity. (2 marks)

(c) Calculate the wind speed and give its direction, either as an angle relative to west or as a bearing. (2 marks)

Question 4 (6 marks)

An aircraft has an airspeed of \(220\text{ km/h}\) and a heading of \(030^\circ\). A steady wind blows due east at \(45\text{ km/h}\).

(a) Calculate the east and north components of the aircraft’s air velocity. (2 marks)

(b) Determine the east and north components of its ground velocity. (2 marks)

(c) Calculate the aircraft’s ground speed and ground-track bearing. (2 marks)

Question 5 (7 marks)

An aircraft must maintain a ground track of bearing \(060^\circ\) at a ground speed of \(180\text{ km/h}\).

The wind has two components: \(25\text{ km/h}\) westward and \(40\text{ km/h}\) southward.

(a) Calculate the east and north components of the required ground velocity. (2 marks)

(b) Determine the east and north components of the aircraft’s required air velocity. (2 marks)

(c) Calculate the required airspeed. (1 mark)

(d) Determine the required heading as a three-digit bearing, to the nearest degree. (2 marks)

Question 6 (8 marks)

An aircraft has an airspeed of \(200\text{ km/h}\), and the wind remains constant throughout the following observations.

On its first leg, the aircraft’s heading is due north. Its ground velocity is measured as \(220\text{ km/h}\) on a bearing of \(010^\circ\).

(a) Determine the wind’s east and north components, its speed, and its direction as a bearing. (5 marks)

The aircraft then turns to a heading due east without changing its airspeed. The wind remains unchanged.

(b) Predict the aircraft’s new ground speed and ground-track bearing. (3 marks)

03Worked solutions and marking guide

These solutions use the suggested mark allocations for this practice paper. Award method marks for correct physics and follow-through marks when a later calculation consistently uses an earlier numerical result. Accept equivalent direction descriptions and reasonable rounding.

Solution 1

(a) Heading is the direction the aircraft points, while ground track is the direction in which it actually moves relative to the ground. Airspeed is the aircraft’s speed relative to the surrounding air, while ground speed is its speed relative to the ground.

The wind can change both the ground track and ground speed without changing the aircraft’s heading or airspeed.

Marking guide (2 marks):

• 1 mark for correctly distinguishing heading from ground track.
• 1 mark for correctly distinguishing airspeed from ground speed.

(b) The ground speed is \(184\text{ km/h}\), and the ground-track bearing is \(013^\circ\).

The aircraft’s air velocity points due north, while the wind contributes an eastward component.

Taking east as positive \(E\) and north as positive \(N\),

\[ \begin{aligned} v_{g,E}&=0+40=40\text{ km/h}\\ v_{g,N}&=180+0=180\text{ km/h} \end{aligned} \]

The ground speed is the magnitude of the ground velocity:

\[ \begin{aligned} v_g&=\sqrt{40^2+180^2}\\ &=\sqrt{34000}\\ &=184.39\text{ km/h}\\ &\approx184\text{ km/h} \end{aligned} \]

The ground-track angle east of north is

\[ \begin{aligned} \theta&=\tan^{-1}\left(\frac{40}{180}\right)\\ &=12.5^\circ \end{aligned} \]

To the nearest degree, the track is \(13^\circ\) east of north, giving a bearing of \(013^\circ\).

Marking guide (4 marks):

• 1 mark for correctly adding the velocity components.
• 1 mark for the ground-speed calculation.
• 1 mark for the track-angle calculation.
• 1 mark for the correct direction and bearing.

Solution 2

(a) The aircraft must head \(14^\circ\) west of north, or approximately \(14.5^\circ\) west of north before rounding.

The wind pushes the aircraft eastward. To maintain a due-north ground track, the aircraft’s air velocity must have an equal westward component.

Let \(\delta\) be the angle west of north. The westward component of the air velocity must have a magnitude of \(50\text{ km/h}\).

\[ \begin{aligned} 200\sin\delta&=50\\ \sin\delta&=\frac{50}{200}\\ \delta&=\sin^{-1}(0.25)\\ &=14.48^\circ \end{aligned} \]

Therefore, the required heading is approximately \(14.5^\circ\) west of north.

Marking guide (3 marks):

• 1 mark for identifying the required westward air-velocity component of \(50\text{ km/h}\).
• 1 mark for setting up the sine relationship correctly.
• 1 mark for the angle and correct side of north.

(b) The required heading is \(346^\circ\).

A bearing is measured clockwise from north. Therefore,

\[ \begin{aligned} \text{Bearing}&=360^\circ-14.48^\circ\\ &=345.52^\circ\\ &\approx346^\circ \end{aligned} \]

Marking guide (1 mark):

• 1 mark for the correct three-digit bearing.

(c) The ground speed is \(194\text{ km/h}\).

Because the aircraft’s westward air-velocity component cancels the eastward wind, the ground velocity has no eastward component. Its northward component determines the ground speed.

Using the magnitude of the air velocity,

\[ \begin{aligned} v_{a,N}&=\sqrt{200^2-50^2}\\ &=\sqrt{37500}\\ &=193.65\text{ km/h} \end{aligned} \]

The eastward components cancel, so the ground speed equals this northward component:

\[ v_g=193.65\text{ km/h}\approx194\text{ km/h} \]

Marking guide (3 marks):

• 1 mark for calculating the northward air-velocity component.
• 1 mark for recognising that the eastward components cancel.
• 1 mark for the correct ground speed.

Solution 3

(a) The ground-velocity components are \(128\text{ km/h}\) east and \(22.6\text{ km/h}\) south.

The ground track is \(10.0^\circ\) south of east, so the east component is positive and the north component is negative.

\[ \begin{aligned} v_{g,E}&=130\cos10.0^\circ\\ &=128.025\text{ km/h}\\ v_{g,N}&=-130\sin10.0^\circ\\ &=-22.573\text{ km/h} \end{aligned} \]

To 3 significant figures, the components are \(+128\text{ km/h}\) east and \(-22.6\text{ km/h}\) north.

Marking guide (2 marks):

• 1 mark for the east component.
• 1 mark for the north component, including its sign.

(b) The wind-velocity components are \(22.0\text{ km/h}\) west and \(22.6\text{ km/h}\) south.

Rearrange the velocity relationship:

\[ \vec v_w=\vec v_g-\vec v_a \]

The aircraft’s air velocity is due east, so its components are \(v_{a,E}=150\text{ km/h}\) and \(v_{a,N}=0\).

Therefore,

\[ \begin{aligned} v_{w,E}&=128.025-150\\ &=-21.975\text{ km/h}\\ v_{w,N}&=-22.573-0\\ &=-22.573\text{ km/h} \end{aligned} \]

Both components are negative, confirming that the wind points towards the southwest.

Marking guide (2 marks):

• 1 mark for using \(\vec v_w=\vec v_g-\vec v_a\).
• 1 mark for the correct wind components.

(c) The wind speed is \(31.5\text{ km/h}\), directed \(46^\circ\) south of west, on a bearing of \(224^\circ\).

Calculate the magnitude of the wind velocity:

\[ \begin{aligned} v_w&=\sqrt{(-21.975)^2+(-22.573)^2}\\ &=31.5\text{ km/h} \end{aligned} \]

The angle south of west is

\[ \begin{aligned} \phi&=\tan^{-1}\left(\frac{22.573}{21.975}\right)\\ &=45.8^\circ \end{aligned} \]

Thus, the wind is directed approximately \(46^\circ\) south of west. Its bearing is \(270^\circ-45.8^\circ\approx224^\circ\).

Marking guide (2 marks):

• 1 mark for the wind-speed calculation.
• 1 mark for the correct direction or bearing.

Solution 4

(a) The air-velocity components are \(110\text{ km/h}\) east and \(191\text{ km/h}\) north.

The heading is \(030^\circ\), meaning \(30^\circ\) clockwise from north.

\[ \begin{aligned} v_{a,E}&=220\sin30^\circ\\ &=110\text{ km/h}\\ v_{a,N}&=220\cos30^\circ\\ &=190.526\text{ km/h} \end{aligned} \]

Marking guide (2 marks):

• 1 mark for the east component.
• 1 mark for the north component.

(b) The ground-velocity components are \(155\text{ km/h}\) east and \(191\text{ km/h}\) north.

The wind has an east component of \(45\text{ km/h}\) and a north component of zero.

\[ \begin{aligned} v_{g,E}&=110+45\\ &=155\text{ km/h}\\ v_{g,N}&=190.526+0\\ &=190.526\text{ km/h} \end{aligned} \]

Marking guide (2 marks):

• 1 mark for the east ground-velocity component.
• 1 mark for the north ground-velocity component.

(c) The ground speed is \(246\text{ km/h}\), and the ground-track bearing is \(039^\circ\).

First, calculate the magnitude:

\[ \begin{aligned} v_g&=\sqrt{155^2+190.526^2}\\ &=245.61\text{ km/h}\\ &\approx246\text{ km/h} \end{aligned} \]

The ground-track angle clockwise from north is

\[ \begin{aligned} \theta&=\tan^{-1}\left(\frac{155}{190.526}\right)\\ &=39.1^\circ \end{aligned} \]

The bearing is therefore \(039^\circ\) to the nearest degree.

Marking guide (2 marks):

• 1 mark for the ground-speed calculation.
• 1 mark for the correct ground-track bearing.

Solution 5

(a) The required ground-velocity components are \(156\text{ km/h}\) east and \(90.0\text{ km/h}\) north.

The ground speed is \(180\text{ km/h}\), and the ground-track bearing is \(060^\circ\).

\[ \begin{aligned} v_{g,E}&=180\sin60^\circ\\ &=155.885\text{ km/h}\\ v_{g,N}&=180\cos60^\circ\\ &=90.0\text{ km/h} \end{aligned} \]

Marking guide (2 marks):

• 1 mark for the east component.
• 1 mark for the north component.

(b) The required air-velocity components are \(181\text{ km/h}\) east and \(130\text{ km/h}\) north.

The wind components are negative in both directions:

\[ v_{w,E}=-25\text{ km/h},\qquad v_{w,N}=-40\text{ km/h} \]

Rearranging the velocity relationship gives

\[ \vec v_a=\vec v_g-\vec v_w \]

Therefore,

\[ \begin{aligned} v_{a,E}&=155.885-(-25)\\ &=180.885\text{ km/h}\\ v_{a,N}&=90.0-(-40)\\ &=130\text{ km/h} \end{aligned} \]

The aircraft needs greater eastward and northward air-velocity components than its required ground-velocity components because the wind pushes it west and south.

Marking guide (2 marks):

• 1 mark for correctly rearranging the velocity relationship.
• 1 mark for the correct air-velocity components.

(c) The required airspeed is \(223\text{ km/h}\).

Airspeed is the magnitude of the air-velocity vector:

\[ \begin{aligned} v_a&=\sqrt{180.885^2+130^2}\\ &=222.75\text{ km/h}\\ &\approx223\text{ km/h} \end{aligned} \]

Marking guide (1 mark):

• 1 mark for the correct airspeed calculation and result.

(d) The required heading is \(054^\circ\).

Both air-velocity components are positive, so the aircraft heads northeast. The heading angle measured clockwise from north is

\[ \begin{aligned} \theta&=\tan^{-1}\left(\frac{180.885}{130}\right)\\ &=54.3^\circ \end{aligned} \]

Therefore, the required heading is \(054^\circ\) to the nearest degree.

Marking guide (2 marks):

• 1 mark for the correct inverse-tangent relationship.
• 1 mark for the correct heading bearing.

Solution 6

(a) The wind speed is \(41.7\text{ km/h}\), directed on a bearing of \(066^\circ\). Its components are \(38.2\text{ km/h}\) east and \(16.7\text{ km/h}\) north.

On the first leg, the ground speed is \(220\text{ km/h}\) and the ground-track bearing is \(010^\circ\).

Calculate the ground-velocity components:

\[ \begin{aligned} v_{g,E}&=220\sin10^\circ\\ &=38.202\text{ km/h}\\ v_{g,N}&=220\cos10^\circ\\ &=216.658\text{ km/h} \end{aligned} \]

The aircraft’s air velocity points due north. Its components are \(0\text{ km/h}\) east and \(200\text{ km/h}\) north.

Subtract the air velocity from the ground velocity to find the wind velocity:

\[ \begin{aligned} v_{w,E}&=38.202-0\\ &=38.202\text{ km/h}\\ v_{w,N}&=216.658-200\\ &=16.658\text{ km/h} \end{aligned} \]

The wind speed is

\[ \begin{aligned} v_w&=\sqrt{38.202^2+16.658^2}\\ &=41.7\text{ km/h} \end{aligned} \]

Its bearing is

\[ \begin{aligned} \theta_w&=\tan^{-1}\left(\frac{38.202}{16.658}\right)\\ &=66.4^\circ\\ &\approx066^\circ \end{aligned} \]

Both components are positive, so the wind blows towards the northeast.

Marking guide (5 marks):

• 2 marks for the correct east and north ground-velocity components.
• 1 mark for correctly subtracting the air velocity to find the wind components.
• 1 mark for the wind-speed calculation.
• 1 mark for the correct wind direction or bearing.

(b) The new ground speed is \(239\text{ km/h}\), and the new ground-track bearing is \(086^\circ\).

On the second leg, the aircraft’s air velocity points due east, so its components are \(200\text{ km/h}\) east and \(0\text{ km/h}\) north.

The wind remains unchanged, with components of \(38.202\text{ km/h}\) east and \(16.658\text{ km/h}\) north.

The new ground-velocity components are

\[ \begin{aligned} v_{g,E}&=200+38.202\\ &=238.202\text{ km/h}\\ v_{g,N}&=0+16.658\\ &=16.658\text{ km/h} \end{aligned} \]

The ground speed is

\[ \begin{aligned} v_g&=\sqrt{238.202^2+16.658^2}\\ &=238.8\text{ km/h}\\ &\approx239\text{ km/h} \end{aligned} \]

The ground-track bearing is

\[ \begin{aligned} \theta_g&=\tan^{-1}\left(\frac{238.202}{16.658}\right)\\ &=86.0^\circ\\ &\approx086^\circ \end{aligned} \]

The aircraft points due east, but its ground track is slightly north of east because the wind has a northward component.

Marking guide (3 marks):

• 1 mark for the correct new ground-velocity components.
• 1 mark for the ground-speed calculation.
• 1 mark for the correct ground-track bearing.

Total: 40 marks.