Plane Speed and Wind Speed: Two-Condition Rate Problem

Question

PROBLEM

Flying to Kampala with tailwind a plane averaged 158 km/h. On the return trip the plane averaged only 112 km/h while flying back into the same wind. Find the speed of the wind and the speed of the plane in still air.

What This Problem Teaches

Setting up systems of equations from real-world scenarios with relative motion
Understanding how wind affects ground speed in opposite directions
Using the elimination method when equations have symmetric structures
Distinguishing between absolute speeds and relative speeds in physics contexts
Recognizing that the same distance constraint creates two related equations

Visualizing the Problem

Flying to Kampala with tailwind a plane averaged 158 km/h. On the return trip the plane averaged only 112 km/h while...

Solution: Method 1 — The Elimination Approach

Step 1 — Define the Variables

Let's call the plane's speed in still air p and the wind speed w (both in km/h).

Step 2 — Set Up the Speed Equations

When flying with a tailwind, the plane's ground speed is the sum of its air speed and wind speed. When flying against a headwind, we subtract the wind speed:

With tailwind: p + w = 158
Against headwind: p - w = 112

Step 3 — Add the Equations to Find Plane Speed

Adding both equations eliminates the wind variable:

(p + w) + (p - w) = 158 + 112
2p = 270
p = 135

Step 4 — Find the Wind Speed

Substitute p = 135 into either original equation. Using the first:

135 + w = 158
w = 23

Solution: Method 2 — The Subtraction Method

Step 1 — Start with the Same System

We have the same two equations:

p + w = 158 ... (1)
p - w = 112 ... (2)

Step 2 — Subtract to Find Wind Speed First

Subtracting equation (2) from equation (1) eliminates the plane speed:

(p + w) - (p - w) = 158 - 112
p + w - p + w = 46
2w = 46
w = 23

Step 3 — Find Plane Speed

Now substitute w = 23 into either equation:

p + 23 = 158
p = 135

The plane's speed in still air is 135 km/h and the wind speed is 23 km/h.

Verification

Let's check our answer by substituting back into both original conditions:

With tailwind: 135 + 23 = 158 km/h ✓
Against headwind: 135 - 23 = 112 km/h ✓

Both conditions are satisfied, confirming our solution is correct.

Watch Out For These

✗ Averaging the Two Speeds

Some students think: "The average of 158 and 112 is 135, so that must be the still-air speed." While this happens to give the right answer here, it's coincidental and doesn't work in general. If the distances were different, this method would fail completely.

✗ Confusing Which Speed is Which

Writing p - w = 158 and p + w = 112. Remember: tailwind helps the plane go faster, so the higher ground speed (158) corresponds to p + w.

✗ Setting Up Distance Equations

Some students try to use distance = rate × time but realize they don't have time or distance information. The key insight is that both trips cover the same distance, so the speed relationships alone are sufficient.

The Pattern Behind This

This problem follows the classic sum-and-difference system pattern:

A + B = sum
A - B = difference

The solution is always:

A = (sum + difference) ÷ 2
B = (sum - difference) ÷ 2

In our case: A = (158 + 112) ÷ 2 = 135 and B = (158 - 112) ÷ 2 = 23.

Important: This shortcut only works when you have one equation that adds two quantities and another that subtracts the same two quantities. The elimination method shown in Method 1 works for all linear systems.

Why This Matters

Wind-speed calculations appear throughout aviation and navigation:

Flight planning: Pilots must account for wind when calculating fuel requirements and arrival times
Weather forecasting: Meteorologists use aircraft reports to measure high-altitude wind speeds
Shipping logistics: Similar calculations apply to boats dealing with currents

What If?

1

Stronger Wind

A plane flies with a tailwind at 170 km/h and returns against the same wind at 100 km/h. Find the plane's still-air speed and the wind speed.

Step 1 — Set up the equations

Let p = plane speed in still air, w = wind speed

p + w = 170 (with tailwind)

p - w = 100 (against headwind)

Step 2 — Add to find plane speed

(p + w) + (p - w) = 170 + 100

2p = 270

p = 135

Step 3 — Find wind speed

135 + w = 170

w = 35

Step 4 — Verify

Check: 135 + 35 = 170 ✓ and 135 - 35 = 100 ✓

Answer: Plane speed = 135 km/h, Wind speed = 35 km/h

2

Find the Distance

A plane's still-air speed is 150 km/h. With a tailwind, it takes 3 hours for a trip. Against the same wind, the return trip takes 5 hours. Find the wind speed and the distance.

Step 1 — Set up using distance = speed × time

Let w = wind speed, d = distance

With tailwind: d = (150 + w) × 3

Against wind: d = (150 - w) × 5

Step 2 — Set distances equal

(150 + w) × 3 = (150 - w) × 5

450 + 3w = 750 - 5w

Step 3 — Solve for wind speed

3w + 5w = 750 - 450

8w = 300

w = 37.5

Step 4 — Find distance

d = (150 + 37.5) × 3 = 187.5 × 3 = 562.5

Answer: Wind speed = 37.5 km/h, Distance = 562.5 km

3

Three-Leg Journey

A plane makes three equal-distance flights: with tailwind at 180 km/h, in still air at 150 km/h, and against headwind at 120 km/h. Are the tailwind and headwind from the same wind system?

Step 1 — Analyze the still-air data

The middle flight gives us the plane's still-air speed directly: p = 150 km/h

Step 2 — Calculate wind speeds

For tailwind: 150 + w₁ = 180, so w₁ = 30 km/h

For headwind: 150 - w₂ = 120, so w₂ = 30 km/h

Step 3 — Compare wind speeds

Since w₁ = w₂ = 30 km/h, the tailwind and headwind have the same magnitude.

Step 4 — Conclusion

Yes, both wind conditions are from the same 30 km/h wind system — the plane flew with it, then without it, then against it.

4

Reverse Engineering

Two planes have still-air speeds of 160 km/h and 140 km/h. In the same wind conditions, their ground speeds are 185 km/h and 165 km/h respectively. Are they both flying with tailwinds, or is one flying against a headwind?

Step 1 — Analyze plane 1

Plane 1: still-air speed = 160 km/h, ground speed = 185 km/h

Wind effect: 185 - 160 = +25 km/h (tailwind)

Step 2 — Analyze plane 2

Plane 2: still-air speed = 140 km/h, ground speed = 165 km/h

Wind effect: 165 - 140 = +25 km/h (tailwind)

Step 3 — Compare wind effects

Both planes experience a +25 km/h wind effect, meaning both have tailwinds.

Step 4 — Conclusion

Both planes are flying with 25 km/h tailwinds in the same direction. Their different ground speeds result from their different still-air capabilities, not different wind conditions.

Frequently Asked Questions

Define variables for the plane's speed in still air (p) and wind speed (w). With a tailwind, the ground speed is p + w. Against a headwind, the ground speed is p - w. Set these equal to the given speeds to create two equations: p + w = 158 and p - w = 112.

Add the two equations to eliminate one variable, then subtract them to eliminate the other. In this problem, adding (p + w = 158) + (p - w = 112) gives 2p = 270, so p = 135. Subtracting gives 2w = 46, so w = 23.

The average of 158 and 112 is 135 km/h, which happens to be correct here, but this is pure coincidence. The method only works when the wind speed is the same in both directions and the distance is the same. For different distances or varying wind conditions, you must solve the system of equations properly.

NJ

Neven Jurkovic, PhD

Professor of Computer Science, Palo Alto College, Alamo Colleges District, San Antonio, TX

Developer of Algebrator

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This solution was prepared with AI assistance and reviewed by Dr. Jurkovic for mathematical accuracy and pedagogical clarity.

2026-06-26