### Video Transcript

An airplane leaves Chicago and makes the 3.00 times 10 to the third kilometer trip to Los Angeles in 18.0 times 10 to the third seconds. A second plane leaves Chicago 1.8 times 10 to the third seconds later and arrives in Los Angeles at the same time. Find the ratio of the average velocities of the two planes. Ignore the curvature of Earth and the difference in altitude between the two cities.

Let’s begin by highlighting some of the important information we’re given. We’re told that the distance between Chicago and Los Angeles is 3.00 times 10 to the third kilometers; we’ll call that value 𝑑. We’re told that the first airplane going to Los Angeles arrives in 18.0 times 10 to the third seconds; we’ll call that time 𝑡 sub one.

Then a second plane leaves 1.8 times 10 to the third seconds later and arrives in Los Angeles at the same time; we’ll call the time it takes the second plane to make the journey 𝑡 sub two, and that’s equal to 18.0 times 10 to the third seconds minus 1.8 times 10 to the third seconds.

We want to solve for the ratio of the average velocities of the two planes. If we call the average velocity of the second plane 𝑣 sub two and the average velocity of the first plane 𝑣 sub one, then we want to solve for their ratio. And just as the problem says, we’ll ignore Earth’s curvature and the fact that the cities are at different altitudes.

Let’s begin our solution by recalling a relationship for average speed. That relationship for average speed says that the average speed of an object 𝑣 is equal to the distance it travels divided by the time it takes that object to travel that distance.

We can apply this relationship to our scenario to solve for 𝑣 two and 𝑣 one. 𝑣 two, the average speed of the second plane, is equal to the distance traveled divided by 𝑡 two. And 𝑣 one, the average speed of the first plane, equals the distance that plane traveled 𝑑 divided by 𝑡 one. If we divide 𝑣 two by 𝑣 one, we get a ratio of 𝑑 divided by 𝑡 two divided by 𝑑 divided by 𝑡 one.

In this fraction, 𝑑, the distance traveled, cancels out, and the fraction simplifies to 𝑡 one divided by 𝑡 two. We can now enter in the values for those two terms. The numerator is 𝑡 one, 18.0 times 10 to the third seconds. And the denominator simplifies to 16.2 times 10 to the third seconds.

Notice that the units cancel out, and we find a ratio of 1.11. That’s the ratio of the average speed of the second plane to the average speed of the first.