a car travels 50kph from point a to b and travels 40kph back. what is the average
Here's the "proof" of this......
Total Distance / Total Time = Average Rate
Let D be the one-way distance....then, the total distance = 2D
And the total time is given by D/R1 + D/R2 ......so we have.....
[2D] / [ D/R1 + D/R2 ] =
[2D] / [D (R2 + R1)/ (R1R2)] =
[2(R1R2)] / (R2 + R1) ........note that this doesn't depend on "D"
This will be the same answer no matter how far it is so to make life easy I am going to say it is 200km
At 50km/h it will take 4 hours
At 40km/hour it will take 5 hours
That is 9 hours altogether to travel 400km
So the average speed it 400/9 km/hour = $$44\frac{4}{9}\;\;km/hour$$
It's always easy to find the average speed in these situations......here's the "formula"...
[2* R1* R2] / [R1 + R2 ] ......where R1 and R2 are the two different rates.......
I didn't know that Chris. Thank you. I'll have to think about it more when I have more time! :)
Here's the "proof" of this......
Total Distance / Total Time = Average Rate
Let D be the one-way distance....then, the total distance = 2D
And the total time is given by D/R1 + D/R2 ......so we have.....
[2D] / [ D/R1 + D/R2 ] =
[2D] / [D (R2 + R1)/ (R1R2)] =
[2(R1R2)] / (R2 + R1) ........note that this doesn't depend on "D"
$$\\(2* R1* R2) / (R1 + R2 ) \\\\
=\dfrac{\frac{2xy\; km^2}{h^2}}{\frac{x\;km}{h}+\frac{y\;km}{h}}\\\\\\
=\dfrac{\frac{2xy\; km^2}{h^2}}{\frac{x+y\;km}{h}}\\\\\\
=\frac{2xy\; km^2}{h^2}\div \frac{x+y\;km}{h}\\\\
=\frac{2xy\; km^2}{h^2}\times \frac{h}{x+y\;km}\\\\
=\frac{2xy\; km}{h}\times \frac{1}{x+y}\\\\
=\frac{2xy\; km}{(x+y)h}\\\\
=\frac{2xy}{(x+y)}\frac{km}{h}\\\\$$
Umm. Not a great deal of help although at least the units are right.
I will have to think about it some more.