# Thread: Average speed of a man walking up a hill.

1. ## Average speed of a man walking up a hill.

Hello everyone, this problem was broadcast on BBC 4's prog 'More Or Less' last Christmas & has been puzzling me ever since:

"A man goes for a walk after his Christmas dinner up a hill behind his house at an average speed of 2 miles/hour. He then returns by exactly the same route at an average speed of 3 miles/hour. What is his overall average speed?"

It is not (2 + 3) / 2. If it is the square root of (2 x 3) = 2.45 mph, could you please explain the vectors involved & why, in spite of the mutually cancelling slopes, it is not simply a horizontal linear average of 2.5 mph.

This was supposed to be done in the head in 5 seconds after a Christmas dinner! Many thanks, Mike. (Aged 66).

2. suppose that the distance he walked uphill is x [miles]. then the time required for uphill travel was x/2 and the units will be in hour. the same in the down hill travel the time will be x/3. then the total distance travelled will be 2x and the total time will be x/2+x/3
divide the first by the second you will find that the average speed will be 2.4 miles/hour

3. ## Reply & thanks to Samer guirguis 2000

Thanks very much Samer for your prompt & sucinct answer to my problem. I really appreciate it, I would never have cracked it for all my messing about. Regards Mike.

4. Originally Posted by nepsesh
Hello everyone, this problem was broadcast on BBC 4's prog 'More Or Less' last Christmas & has been puzzling me ever since:

"A man goes for a walk after his Christmas dinner up a hill behind his house at an average speed of 2 miles/hour. He then returns by exactly the same route at an average speed of 3 miles/hour. What is his overall average speed?"

It is not (2 + 3) / 2. If it is the square root of (2 x 3) = 2.45 mph, could you please explain the vectors involved & why, in spite of the mutually cancelling slopes, it is not simply a horizontal linear average of 2.5 mph.

This was supposed to be done in the head in 5 seconds after a Christmas dinner! Many thanks, Mike. (Aged 66).
Let x denote the distance of one trip. Then the complete distance is 2x.

According to the definition of speed you'll get:

$time=\dfrac{distance}{speed}$ . Therefore the time for both trips is:

$\dfrac x2 + \dfrac x3=t$

The overall average speed is:

$v_{average} = \dfrac{2x}{\frac x2 + \frac x3}=\dfrac{2x}{\frac{5x}6} = \dfrac{12}5=2.4$

EDIT: Too late again ...