Airplane Speed A man flies a small airplane from Fargo to...

Airplane Speed A man flies a small airplane from Fargo to Bismarck, North Dakota-a distance of 180 mi. Because he is flying into a headwind, the trip takes him 2 h. On the way back, the wind is still blowing at the same speed, so the return rip takes only 1 $\mathrm{h} 12$ min. What is his speed in still air, and how fast is the wind blowing?

Airplane Speed A man flies a small airplane from Fargo to Bismarck, North Dakota-a distance of 180 mi. Because he is flying into a headwind, the trip takes him 2 h. On the way
back, the wind is still blowing at the same speed, so the return rip takes only 1 $\mathrm{h} 12$ min. What is his speed in still air, and how fast is the wind blowing?

Key Concepts

Relative Motion

Relative motion involves understanding how the movement of an object is observed differently when influenced by an external factor, like wind. In the context of flying, the effective speed of the airplane relative to the ground is a combination of its speed in still air and the wind’s speed, with the wind either adding to or subtracting from the airplane’s speed depending on its direction.

Headwind and Tailwind

A headwind is a wind blowing against the direction of travel, which reduces the airplane's effective ground speed, while a tailwind blows in the same direction of travel, increasing the effective ground speed. Recognizing these conditions is essential in analyzing how external forces like wind affect travel time and speed.

Speed, Distance, and Time Relationship

This fundamental concept states that distance equals speed multiplied by time. It is crucial for problems like this, where known distances and travel times are used to establish relationships between the airplane's speed in still air and the wind speed.

Systems of Equations

When multiple scenarios with the same unknowns are presented, such as flying with and against the wind, a system of linear equations can be set up. These equations are solved simultaneously to determine the airplane’s speed in still air and the wind speed, making it a vital algebraic method in solving applied problems.

Transcript

00:01 Okay, so in this problem, it's actually a distance equals rate times time problem.

00:05 So we'll have d equals rt, but we have two different rates here.

00:14 And so we have the rate of the plane.

00:17 So i'll go ahead and call it speed.

00:19 It's really right, but we'll go speed of plane, and then y will be our speed of wind.

00:31 And so for this first problem, or the first linear equation, we have 180 equals, and so we're going against the wind.

00:45 So we'll have the speed of the plane minus the speed of the wind.

00:49 And it's going to take us a total of two hours.

00:52 And we're going to go ahead and convert that into minutes because our next one is going to be weird if we just try and keep it in hours.

00:58 So we'll have 120.

01:01 And our next one, we're going to have the same day.

01:04 So it's going to be 180, but in this case we're going with the wind, so we'll have the speed of the plane plus the speed of the wind.

01:11 But this time we will also, we'll be doing it in just an hour and 12 minutes, which converts to 72 minutes.

01:20 So now that we've got that through, let's go ahead and clean this up a little bit.

01:25 So this first equation, so this is our system of linear equations here.

01:32 But i want to go ahead and simplify this.

01:34 And so the first thing i'm going to say is this first one can be divisible, is divisible by 60.

01:40 So then if we divide that by 60, we'll end up getting three.

01:48 We'll have three equals, yeah, three equals x minus y times two...

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