# Distance

• Mar 17th 2010, 08:38 PM
MATNTRNG
Distance
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?
• Mar 18th 2010, 07:27 AM
Hello MATNTRNG
Quote:

Originally Posted by 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 $t$ represents the elapsed time in hours, and $s$ 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:
$\frac{\text{distance from starting point}}{\text{time taken}}=\frac{FH}{OH}$
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 $t_1$ hours elapses before Bob drops Chris, and returns to pick up Peter, taking a further time $t_2$ to meet him. These times are represented by the distances $OA$ and $AB$ on the graph. Then, using $s = vt$, we have:
$AC = ct_1,\; EC = ct_2,\; BD = p(t_1+t_2)$
Now:
$AC = AE + EC$
$=BD + EC$
$\Rightarrow ct_1 = p(t_1+t_2) +ct_2$

$\Rightarrow t_2 = \frac{c-p}{c+p}\cdot t_1$
...(1)
The second part of this first cycle is where Bob and Peter overtake Chris, represented by the line-segment $DF$. Now since the car travels at a constant speed, and the two walkers walk at the same speed, $OC \parallel DF$ and $OD \parallel CF$. $ODFC$ is therefore a parallelogram, and hence:
$CF = OD$ and $AC = DF$
and therefore:
$CG = OB$ and $BD = GF$

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

$=OA + OB$

$=t_1+(t_1+t_2)$

$=2t_1+t_2$

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

$=\frac{3c+p}{c+p}\cdot t_1$
...(2)
Also:
$FH = FG+GH$
$=DB + AC$

$=p(t_1+t_2)+ct_1$

$=(p+c)t_1+pt_2$

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

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

$=\frac{(3p+c)c}{c+p}\cdot t_1$
...(3)
So, from (2) and (3):
$\frac{FH}{OH}=\frac{(3p+c)c}{3c+p}$
which is independent of $t_1$, 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).