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

 Sep 9, 2014

Best Answer 

 #5
avatar+98173 
+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/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"

 

 Sep 9, 2014
 #1
avatar+4472 
0

Please post the complete question.

 Sep 9, 2014
 #2
avatar+99352 
+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 =    $$44\frac{4}{9}\;\;km/hour$$  

 Sep 9, 2014
 #3
avatar+98173 
+5

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.......

 

 Sep 9, 2014
 #4
avatar+99352 
0

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

 Sep 9, 2014
 #5
avatar+98173 
+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/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"

 

CPhill Sep 9, 2014
 #6
avatar+99352 
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.  

 Sep 9, 2014
 #7
avatar+99352 
0

Yep got it.  Thanks Chris.   

 Sep 9, 2014

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