Hello MATNTRNG Quote:

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**MATNTRNG** 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 $\displaystyle t$ represents the elapsed time in hours, and $\displaystyle s$ the distance from the starting point in miles.

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:$\displaystyle \frac{\text{distance from starting point}}{\text{time taken}}=\frac{FH}{OH}$

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 $\displaystyle t_1$ hours elapses before Bob drops Chris, and returns to pick up Peter, taking a further time $\displaystyle t_2$ to meet him. These times are represented by the distances $\displaystyle OA$ and $\displaystyle AB$ on the graph. Then, using $\displaystyle s = vt$, we have:$\displaystyle AC = ct_1,\; EC = ct_2,\; BD = p(t_1+t_2)$

Now:$\displaystyle AC = AE + EC$$\displaystyle =BD + EC$

$\displaystyle \Rightarrow ct_1 = p(t_1+t_2) +ct_2$

$\displaystyle \Rightarrow t_2 = \frac{c-p}{c+p}\cdot t_1$ ...(1)

The second part of this first cycle is where Bob and Peter overtake Chris, represented by the line-segment $\displaystyle DF$. Now since the car travels at a constant speed, and the two walkers walk at the same speed, $\displaystyle OC \parallel DF$ and $\displaystyle OD \parallel CF$. $\displaystyle ODFC$ is therefore a parallelogram, and hence:$\displaystyle CF = OD$ and $\displaystyle AC = DF$

and therefore:$\displaystyle CG = OB$ and $\displaystyle BD = GF$

$\displaystyle \Rightarrow OH = OA + AH$$\displaystyle =OA + CG$

$\displaystyle =OA + OB$

$\displaystyle =t_1+(t_1+t_2)$

$\displaystyle =2t_1+t_2$

$\displaystyle =\left(2+\frac{c-p}{c+p}\right)t_1$, from (1)

$\displaystyle =\frac{3c+p}{c+p}\cdot t_1$ ...(2)

Also:$\displaystyle FH = FG+GH$$\displaystyle =DB + AC$

$\displaystyle =p(t_1+t_2)+ct_1$

$\displaystyle =(p+c)t_1+pt_2$

$\displaystyle =\left(p+c+\frac{p(c-p)}{c+p}\right)t_1$, from (1)

$\displaystyle =\left(\frac{p^2+2pc+c^2+pc-p^2}{c+p}\right)t_1$

$\displaystyle =\frac{(3p+c)c}{c+p}\cdot t_1$ ...(3)

So, from (2) and (3):

$\displaystyle \frac{FH}{OH}=\frac{(3p+c)c}{3c+p}$

which is independent of $\displaystyle t_1$, 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).

Grandad