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). (Worried)

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

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. (Nod)

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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). (Worried)

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Let x denote the distance of one trip. Then the complete distance is 2x.

According to the definition of speed you'll get:

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

$\displaystyle \dfrac x2 + \dfrac x3=t$

The overall average speed is:

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

EDIT: Too late again ...(Crying)