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a car travels 50kph from point a to b and travels 40kph back. what is the average

Guest Sep 9, 2014

#5**+10 **

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/R_{1} + D/R_{2} ......so we have.....

[2D] / [ D/R_{1} + D/R_{2} ] =

[2D] / [D (R_{2} + R_{1})/ (R_{1}R_{2})] =

[2(R_{1}R_{2})] / (R_{2} + R_{1}) ........note that this doesn't depend on "D"

CPhill Sep 9, 2014

#2**+5 **

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 = $$44\frac{4}{9}\;\;km/hour$$

Melody Sep 9, 2014

#3**+5 **

It's always easy to find the average speed in these situations......here's the "formula"...

[2* R_{1}* R_{2}] / [R_{1} + R_{2} ] ......where R_{1} and R_{2} are the two different rates.......

CPhill Sep 9, 2014

#4**0 **

I didn't know that Chris. Thank you. I'll have to think about it more when I have more time! :)

Melody Sep 9, 2014

#5**+10 **

Best Answer

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/R_{1} + D/R_{2} ......so we have.....

[2D] / [ D/R_{1} + D/R_{2} ] =

[2D] / [D (R_{2} + R_{1})/ (R_{1}R_{2})] =

[2(R_{1}R_{2})] / (R_{2} + R_{1}) ........note that this doesn't depend on "D"

CPhill Sep 9, 2014

#6**0 **

$$\\(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.

Melody Sep 9, 2014