C. Geometry and Algebra of Vectors · Vectors in Two-Space and Three-Space

Applications: Velocity, Wind and Current

C2.3About 75 minutes

Learning goals

I can explain heading, track, airspeed and ground speed, and translate wind directions.
I can find a ground velocity from an air (or water) velocity and a wind (or current).
I can find the heading needed to follow a given track.
I can solve river-crossing problems, including crossing time and drift, and velocity problems in three-space.

Words to know

heading
The direction a plane or boat is pointed.

track (course)
The direction a craft actually moves over the ground.

airspeed
The speed of a plane relative to the air.

ground speed
The speed of a craft relative to the ground: the magnitude of the ground velocity.

relative velocity
The velocity of an object as seen from a moving reference, such as the air, the water or another car.

Where you point is not where you go

A pilot points a plane north-east, but a crosswind pushes it sideways. A canoeist points straight across the Ottawa River, but the current carries the canoe downstream. In both cases the actual motion is the vector sum of two velocities. This is the most famous application of vector addition, and it is the curriculum's own sample problem for this unit.

The language of navigation

Heading: the direction the craft is pointed.
Airspeed (or speed relative to the water): how fast the craft moves through the air (or water).
Air velocity (or water velocity) 𝐚: airspeed together with heading.
Wind velocity (or current) 𝐰: how the air (or water) itself moves. Wind is reported by the direction it blows from: "wind from the south" blows toward the north. Currents are the other way round: they are named by the direction they flow toward, so "a current of 3 km/h toward 120°" moves the water on a bearing of 120°.
Ground velocity 𝐠: the actual motion over the ground. Its magnitude is the ground speed and its direction is the track (or course).

The key equation is

𝐠 = 𝐚 + 𝐰 (ground velocity = air velocity + wind velocity)

You can solve it with a triangle (cosine and sine laws) or with components. Always draw the vector triangle with north lines first.

Worked example

Example 1: The curriculum's sample problem

A plane on a heading of N 27° E has an air speed of 375 km/h. The wind is blowing from the south at 62 km/h. Determine the actual direction of travel of the plane and its ground speed.

Wind from the south blows toward the north: 𝐰 = 62 km/h [N].
Draw 𝐚 (375 km/h at N 27° E), then 𝐰 tip to tail. The angle between 𝐚 and 𝐰 (tail to tail) is 27°, so the angle inside the triangle is 180° − 27° = 153°.
Cosine law: |𝐠|² = 375² + 62² − 2(375)(62) cos 153° ≈ 140 625 + 3844 + 41 432 ≈ 185 901, so |𝐠| ≈ 431.2 km/h.
Sine law for the angle α between 𝐠 and 𝐚: sin α = 62 sin 153° / 431.16 ≈ 0.0653, so α ≈ 3.7°.
The wind pushes the plane toward north, so the track is 27° − 3.7° = 23.3° east of north.
Check with components (x east, y north): 𝐚 = [375 sin 27°, 375 cos 27°] ≈ [170.25, 334.13]; 𝐠 ≈ [170.25, 396.13]; |𝐠| ≈ 431.2 and tan⁻¹(170.25 ÷ 396.13) ≈ 23.3° from north ✓

Ground speed ≈ 431.2 km/h, actual direction ≈ N 23.3° E.

The plane points N 27° E, but the south wind adds a northward push, so the ground velocity is a little faster and points a little more toward north (N 23.3° E).

Crossing a river

For a boat, 𝐠 = 𝐛 + 𝐜: the ground velocity is the boat's velocity relative to the water plus the current. Two classic questions:

Head straight across. The current carries you downstream, but the crossing time depends only on the across-the-river part of your velocity: time = width ÷ (speed straight across).
Land directly opposite. Aim partly upstream so that the upstream part of your velocity cancels the current. Then only the remaining part carries you across, so the crossing takes longer.

Worked example

Example 2: A canoe on a river

A canoeist can paddle at 4 km/h in still water. The river is 200 m wide and flows at 3 km/h. (a) If she heads straight across, how long does it take, how far downstream does she land, and what is her ground velocity? (b) In what direction should she head to land directly opposite, and how long will that take?

(a) Across speed = 4 km/h. Time = 0.2 km ÷ 4 km/h = 0.05 h = 3 min.
Downstream drift = 3 km/h × 0.05 h = 0.15 km = 150 m.
Ground velocity: √(4² + 3²) = 5 km/h, at tan⁻¹(3/4) ≈ 36.9° downstream from straight across.
(b) The upstream part of her velocity must cancel the current: 4 sin θ = 3, so sin θ = 0.75 and θ ≈ 48.6° upstream from straight across.
Her speed straight across is then 4 cos θ = √(16 − 9) = √7 ≈ 2.65 km/h.
Time = 0.2 ÷ 2.6458 ≈ 0.0756 h ≈ 4.5 min.

(a) 3 min, lands 150 m downstream, ground velocity 5 km/h at 36.9° downstream from straight across. (b) Head ≈ 48.6° upstream from straight across; crossing takes ≈ 4.5 min.

Left: heading straight across gives the fastest crossing but lands downstream. Right: aiming 48.6° upstream cancels the current, so the canoe lands directly opposite but crosses more slowly.

Worked example

Example 3: Finding the heading

A pilot wants to fly due east (track 090°). The plane's airspeed is 500 km/h and the wind blows from 330° at 80 km/h. Find the heading the pilot must fly and the ground speed.

Wind from 330° blows toward 150°. With x east and y north: 𝐰 = [80 sin 150°, 80 cos 150°] ≈ [40.00, −69.28].
The ground velocity points due east: 𝐠 = [g, 0]. Then 𝐚 = 𝐠 − 𝐰 = [g − 40.00, 69.28].
|𝐚| = 500: (g − 40.00)² + 69.28² = 500², so (g − 40.00)² = 250 000 − 4800 = 245 200 and g − 40.00 ≈ 495.18 (positive, since the plane flies east). So g ≈ 535.2.
Heading of 𝐚 = [495.18, 69.28]: tan⁻¹(69.28 ÷ 495.18) ≈ 8.0° north of east, so the bearing is 90° − 8.0° = 082.0°.
Makes sense: the wind pushes south-east, so the pilot aims a little north of east, and the eastward part of the wind adds to the speed.

Heading ≈ 082.0°, ground speed ≈ 535.2 km/h.

To stay on a due-east track, the pilot heads 8° north of east. The wind (toward 150°) then bends the path back to due east.

Worked example

Example 4: A drone in three-space

A delivery drone's velocity relative to the air is [4, 3, 1] m/s (x east, y north, z up). A wind of [−2, 1, 0] m/s is blowing. Find the drone's velocity relative to the ground and its speed.

𝐠 = 𝐚 + 𝐰 = [4 − 2, 3 + 1, 1 + 0] = [2, 4, 1].
Speed = √(4 + 16 + 1) = √21 ≈ 4.58 m/s.

[2, 4, 1] m/s, speed ≈ 4.58 m/s.

Checklist for velocity problems

Translate wind directions: from X means toward the opposite direction.
Write 𝐠 = 𝐚 + 𝐰 (or 𝐠 = 𝐛 + 𝐜) and mark which vector is unknown.
Sketch the triangle with a north line at every vertex.
Solve with the cosine and sine laws or with components; components are safest when several vectors are involved.
Answer in the form asked: ground speed and track, or heading and airspeed.

Just for fun

A pilot's joke: "Time flies like an arrow. Airplanes fly like vector sums."

Hands-on activity

The moving-paper river

You need: a sheet of paper (a long strip works best), a ruler, a pencil, a table, and a partner.

Tape the ruler to the table so it points straight away from you. This is the "straight across" direction over the ground.
Lay the paper under the ruler's edge. Your partner will slide the paper slowly and steadily to the right, about 2 cm each second (count "one thousand one…"). The paper is the moving water.
While the paper moves, draw a line along the ruler's edge at about 3 cm each second, so your pencil moves straight across over the table (the ground).
Lift the ruler and look at the line on the paper. It shows your path relative to the water.
Measure the angle between the drawn line and the straight-across direction on the paper.

What to look for: the line on the paper slants against the paper's motion (toward the left). That's the heading a boat needs to go straight across: velocity relative to water = ground velocity − current. With 3 cm/s across and 2 cm/s of "current", the slant is about tan⁻¹(2/3) ≈ 34° from straight across. Your result will vary because hand speeds are not perfectly steady.

Your turn

A plane heads north at an airspeed of 300 km/h. The wind blows from the west at 40 km/h. Find the ground velocity.

A motorboat heads north (straight across a river) at 12 km/h relative to the water. The current flows east at 5 km/h. Find the ground velocity.

For the boat above, the river is 300 m wide. How long does the crossing take and how far downstream does the boat land?

In what direction should the boat above head to land directly opposite its start, and what is its speed across?

Explain the difference between a plane's heading and its track.

A drone's air velocity is [5, −2, 3] m/s and the wind is [1, 4, −1] m/s. Find the ground velocity and speed.

Write a wind "from the north-east at 30 km/h" in components (x east, y north).

More practice: warm-up

A wind "from the west" blows toward which direction?

A plane with airspeed 250 km/h has a 50 km/h tailwind (wind in the same direction). Ground speed?

The same plane flies into a 50 km/h headwind. Ground speed?

Complete: ground velocity = air velocity + ___.

Is ground speed a scalar or a vector?

More practice: on your own

A plane flies at airspeed 400 km/h on a heading of 060°. The wind blows from 180° at 50 km/h. Find the ground velocity.

A swimmer swims straight across a 60 m river at 1.5 m/s relative to the water. The current is 0.8 m/s. Find the crossing time, the drift and the swimmer's speed over the ground.

A plane must fly due north at an airspeed of 250 km/h. The wind blows from the east at 50 km/h. Find the heading and the ground speed.

A kayak heads on a bearing of 040° at 8 km/h relative to the water, in a current of 3 km/h toward 120°. Find its ground velocity.

Car A drives at 80 km/h [N] and car B at 60 km/h [E]. Find the velocity of B relative to A (that is, 𝐯_B − 𝐯_A).

Challenge

A pilot wants to fly due south at airspeed 320 km/h. The wind blows from 270° at 60 km/h. Find the heading and the ground speed.

A plane flies on a heading of 045° at an airspeed of 500 km/h. After 2 h it is 900 km from its start, on a bearing of 050°. Find the wind velocity.

A city is 600 km from an airport on a bearing of 300°. A plane has an airspeed of 450 km/h and the wind blows from 030° at 60 km/h. Find the heading the pilot should fly and the flight time.

Check your understanding

Quiz

The wind is "from the south at 62 km/h". Which vector describes it?

Which equation relates the velocities of a plane?

A boat heads straight across a 100 m river at 2 m/s while the current flows at 1.5 m/s. How long does it take to cross?

A plane heads north at 200 km/h in a wind blowing toward the east at 50 km/h. What is its ground speed?

To land directly opposite its start, a boat crossing a river should head…

A drone's air velocity is [3, 0, 4] m/s and there is no wind. What is its ground speed?

For parents and teachers

Big idea (C2.3): a plane's or boat's real motion is the vector sum of its own velocity through the air or water and the velocity of the air or water: 𝐠 = 𝐚 + 𝐰.

How to help: The moving-paper activity takes five minutes with two people and makes "relative velocity" visible. When checking answers, ask: "Did the wind push the plane the way your answer says?"

Common difficulties:

Treating "wind from the west" as blowing west (it blows east).
Adding speeds instead of velocities (only correct for a pure headwind or tailwind).
Confusing heading (where it points) with track (where it goes).
In river problems, using the ground speed instead of the across-the-river speed to find the crossing time.