Bob, Peter, and Chris travel together. Peter and Chris are good hikers; each walks p miles per hour. Bob has a bad foot and drives a small car in which two people can ride, but not three; the car covers c miles per hour. The three friend adopted the following scheme: They start together. Chris rides in the car with Bob and Peter walks. After a while, Bob drops Chris who walks on; Bob returns to pick up Peter, and then Bob and Peter ride in the car until they overtake Chris. At this point, they change. Chris rides and Peter walks just as they started and the whole procedure is repeated as often as necessary. How far did they go after an hour?
Study the attached distance-time graph, where represents the elapsed time in hours, and the distance from the starting point in miles.
Originally Posted by MATNTRNG
The red line represents the distance of the car from the starting-point, the blue lines, that of the walkers. The graph shows one complete cycle of the scheme, whereby the three friends are all together again, ready to begin another cycle.
We need to find the overall speed in mph from the beginning to the end of this cycle; that is, the fraction:Any further cycles will have exactly the same average speed (as we shall see) and therefore this fraction will represent how far they went in one hour.
Suppose that a time hours elapses before Bob drops Chris, and returns to pick up Peter, taking a further time to meet him. These times are represented by the distances and on the graph. Then, using , we have:Now:The second part of this first cycle is where Bob and Peter overtake Chris, represented by the line-segment . Now since the car travels at a constant speed, and the two walkers walk at the same speed, and . is therefore a parallelogram, and hence:
and and therefore:Also:So, from (2) and (3):
which is independent of , and hence represents the average speed over any complete cycle. Hence it is the number of miles travelled in one hour (assuming that at the end of the hour the three friends are at the same point).