SOLUTION: Joe and Frank want to meet each other half way between their cities. The distance between their towns is 36 miles. Both travel at6 miles per hour. Joe takes a carrier pigeon and se

Question 1200144: Joe and Frank want to meet each other half way between their cities. The distance between their towns is 36 miles. Both travel at6 miles per hour. Joe takes a carrier pigeon and sets it off toward Frank. The pigeon travels at 18 miles an hour. When it reaches Frank it turns around immediately and returns to Joe. What is the distance the pigeon covers when the two friends meet. The pigeon does not take a rest.
Found 3 solutions by Theo, ikleyn, math_tutor2020:
Answer by Theo(13342)   (Show Source): You can put this solution on YOUR website!
joe and frank both travel at 6 miles per hour.
the pigeon travels at 18 miles per hour.

rate * time = distance

the pigeon and frank will meet when both have traveled the same amount of time.

equation for frank is 6 * T = x
equation for the pigeon is 18 * T = 36 - x
solve for x in both equation to get x = 6 * T and x = 36 - 18 * T
that means that 6 * T = 36 - 18 * T
add 18 * T to both sides of the equation to get 24 * T = 36
solve for T to get T = 1.5 hours.

the pigeon and frank will meet when both have traveled for 1.5 hours.
frank will have traveled 9 miles.
the pigeon will have traveled 27 miles.

in 1.5 hours both frank and joe have traveled 9 miles towards each other.
this means they are still 18 miles distance from each other.

the pigeon is now traveling back towards joe at 18 miles an hour.
joe is traveling at 6 miles per hour toward the pigeon.
they will meet when both have traveled the same amount of time.

the equation for joe is 6 * T = x
the equation for the pigeon is 18 * T = 18 - x
solve for x in both equations to get:
x = 6 * T and x = 18 - 18 * T
you get 6 * T = 18 - 18 * T
add 18 * T to both sides of the eqution to get 24 * T = 18
solve for T to get T = 18/24 = .75 hours.

the pigeon and joe will meet in .75 hours.
joe will have traveled 6 * .75 = 4.5 miles.
the pigeon will have traveled 18 * .75 = 13.5 miles.

the pigeon traveled 27 miles to get to frank and then traveled another 13.5 miles to get back to joe.
total air time distance traveled by the pigeon is therefore 27 miles plus 13.5 miles = 40.5 miles.

the total distance traveled by the pigeon while the pigeon was in the air and while the pigeon was with joe and frank would be calculated as follows:

27 miles in 1.5 hours to fly to frank.
13.5 miles in .75 hours to fly back to joe.
total flying time is 40.5 hours.
since it takes 3 hours for frank to meet joe, then there is an additional .75 hours that the pigeon is with joe while joe is walking towards frank.
at 6 miles per hour, that would be another 4.5 miles that the pigeon is traveling with joe.
total distance the pigeon has traveled is therefore 40.5 + 4.5 = 45 miles.
40.5 miles while flying and 4.5 miles while traveling with joe.

this is what i think is your solution, based on my understanding of the problem.
theo

Answer by ikleyn(52756)   (Show Source): You can put this solution on YOUR website!
.
Joe and Frank want to meet each other half way between their cities. The distance between their towns is 36 miles.
Both travel at6 miles per hour. Joe takes a carrier pigeon and sets it off toward Frank.
The pigeon travels at 18 miles an hour. When it reaches Frank it turns around immediately and returns to Joe.
What is the distance the pigeon covers when the two friends meet. The pigeon does not take a rest.
~~~~~~~~~~~~~~~~~~~

            It is well known entertainment problem.
            Its standard solution is in 3 lines.

Joe and Frank approach each other at the rate 6+6 = 12 miles per hour. Hence, they will meet each other in 36/12 = 3 hours. During these 3 hours, flying at the rate of 18 miles per hour, the pigeon will cover the distance of 3*18 = 54 miles. ANSWER

                            Solved.

////////////////

For the solution, it does not matter that their rates are numerically equal, as it is given in the problem.

The quantities that really matter are

                (a) the time until they meet each other,

        and

                (b) the rate of flying of the pigeon.

\\\\\\\\\\\\\\\\\\\

Hello, after looking the solutions from @Theo and from @math_tutor2020,
I can tell you that they both are incorrect and irrelevant.

Simply both these tutors misread the task.

So, ignore their posts - for peace in your mind.

If in the future you will get other solutions with different answer than my, ignore them, too.

Answer by math_tutor2020(3816)   (Show Source): You can put this solution on YOUR website!

Draw a number line. Plot A and B on it
A = 0
B = 36
Then plot C at the midpoint 18

I'll have Joe start at point A, and Frank start at point B.
Both men are traveling to meet at point C.

The pigeon starts at point A.
Its speed is 18 mph, so it travels a distance of 18t where t is the elapsed time in hours.
distance = rate*time

Meanwhile, Frank travels a distance of 6t miles.
Add the expressions 18t and 6t. Then set the sum equal to the total distance 36 to solve for t.
This is because the pigeon meets Frank, so the two must travel a total distance of 36 miles.

18t+6t = 36
24t = 36
t = 36/24
t = 1.5

The pigeon meets Frank at exactly the 1.5 hour marker.

The pigeon has traveled 18t = 18*1.5 = 27 miles in this time frame.
Joe has traveled 6t = 6*1.5 = 9 miles in this time frame.
This means the pigeon would travel 27-9 = 18 additional more miles to get back to Joe if Joe stopped moving.

However, Joe keeps moving.
His location on the number line is still 6t.

When flying back to Joe, the pigeon's location on the number line is 27-18(t-1.5)
The 27 refers to how far the pigeon has flown to the right. Then we subtract off 18(t-1.5) to go to the left.
The t-1.5 offsets the time value (to account for the fact the bird has flown 1.5 hours already)

To summarize:
Joe's location is 6t
The pigeon's location after 1.5 hours is 27-18(t-1.5)

Set them equal to each other to determine the time value when the pigeon meets Joe again.
27-18(t-1.5) = 6t
27-18t+27 = 6t
54-18t = 6t
54 = 6t+18t
54 = 24t
t = 54/24
t = 9/4
t = 2.25

The pigeon meets Joe again at exactly the 2.25 hour marker.
The pigeon has flown a total distance of 18t = 18*2.25 = 40.5 miles.

Check out this interactive GeoGebra applet.
https://www.geogebra.org/m/uubrrczn
If the page is blank, then hover your mouse over it to have the curved arrow show up.
Then click on the arrow. The page should refresh to show the interactive diagram.

Move the slider t so that you can see Joe, Frank, and the pigeon move.
P1 represents the pigeon's location in the interval 0 < t < 1.5
P2 represents the pigeon's location in the interval 1.5 < t < 3
The points J and P2 line up perfectly when t = 2.25

Answer: 40.5 miles

Edit: I'm realizing I misread the instructions a bit. The answer of 40.5 miles would be correct if it asked "What is the distance the pigeon covers when the pigeon gets back to Joe".
Refer to the solution the tutor @ikleyn has written for the correct answer of 54 miles.

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