Does this even make sense?

Question

At. Louis and Portland are 2060 miles apart. A small plane leaves Portland traveling toward St. Louis at an average of 90mph. Another plane leaves St. Louis at the same time, traveling toward Portland averaging 116mph. How long will it take them to meet? 

Answer 1

I think an easier way to solve this problem is to let x = number of hours the planes traveled. They started and met at the same time so they travel for the same amount of time. They fly toward each other, one plane covering part of the 2060 miles and the other plane covering the other part. Use this fact to get an equation where the sum of their distances traveled equals 2060 miles. Use distance = rate x time, to get the expression 90x for the distance the first plane travels and 116x for the second.

 

So the equation is

 

90x + 116x = 2060

 

206x = 2060

 

x =10

 

So the planes fly 10 hours until they meet. Hopefully, they just pass by at a safe distance as they go in opposite directions.

Answer 2

Not withstanding Kehley's beautiful solution, this type of problem requires a higher math. First of all, the distance between the two cities is measured on the ground but the distance the airplanes will fly is longer because they will be at altitude. Since the two cities are far apart we have to take into account the curvature of the earth. The two cities are on an arc but the airplanes fly a longer outside arc.

 

The computation of an average speed is interesting, too. If the plane takes off at zero speed then the longer it flies the greater its average speed will be over time. So, you couldn't actually use Kehley's solution because these factors would be omitted. Another subtle factor is that planes flying east fly at one altitude while planes flying west fly either 1000 ft above or below so they don't in fact run into each other.

 

Wind is an extremely mitigating factor. Suppose the problem said there was a cross wind at 40 knots 40 degrees off the flight path of the east bound plane.

 

Yet another problem is the fact that these are small planes. Small planes usually hold about 40-50 gallons of fuel. The burn rate for a single engine prop plane is about 10 gal/hour (I'm a pilot.) Therefore, every 4 hours or so the pilots will need to land and refuel. That will take about an hour because you have to slow down and descend, get in the pattern, land, find the fuel pump, pump the fuel, pay for the fuel, restart the engine, get clearance and then take off back to altitude. 

 

Better to stick with trains where the trains will actually meet.

 

 

Answer 3

Hi Steve,

This does make sense. It's an aviation version of the two trains that are bound for one another. Basically, we need to determine how many hours it will take for them to reach the same point. IF they were traveling at the same speed, they would meet at the midpoint. BUT since that's not the case, we have to figure out where they will meet and how long it will take one of the planes to reach that point (it shouldn't matter which one you choose, since they'll both be at that point at the same time).

 

So, if the total distance is 2060mi then one plane will travel x miles and the other will travel 2060-x miles. It doesn't matter which plane is which at this point.

Picking which plane travels x miles, we can assign it a speed (let's say it's the 90mph plane). This plane is in the air for T=x/90 hours (recall that the units are miles / mph, so the miles cancel out and hours becomes the numerator).

The other plane, which travels 2060-x miles, moves at the other speed (in this case, 116mph). This plane is in the air for T=(2060-x)/116 hours.

Since we're trying to figure out where they'll meet, we know the time in the air is the same. This means we can set each plane equal to the other, such that: x/90=(2060-x)/116.

Now we can solve for x, and then find how long the planes are in the air by solving for T.