Question 930656: if you travel the same way going and coming back... you go 40 mph going there and 60 mph coming back, how many mph do you average (not 50 mph).
Found 3 solutions by Alan3354, TimothyLamb, MathTherapy:
Answer by Alan3354(69443)   (Show Source):
You can put this solution on YOUR website! if you travel the same way going and coming back... you go 40 mph going there and 60 mph coming back, how many mhp do you average (not 50 mph).
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Avg RT speed = 2*40*60/(40+60)
= 48 mi/hr
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It's less than the average of the 2 speeds because more time is spent going 40 mi/hr.
The distance is not relevant.
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It's the basis of the Mickelson-Morley experiment in the mid 19th century that showed no effects of the "ether," and the discarding of the ether theory.
Answer by TimothyLamb(4379)  (Show Source):
You can put this solution on YOUR website! s = d/t
t = d/s
x = time go
x = d/40
y = time return
y = d/60
total time:
x + y = d/40 + d/60
avg speed = d/( d/40 + d/60 )
avg speed = d/d( 1/40 + 1/60 )
avg speed = 1/( 1/40 + 1/60 )
avg speed = 24 mph
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Answer by MathTherapy(10555)  (Show Source):
You can put this solution on YOUR website!
if you travel the same way going and coming back... you go 40 mph going there and 60 mph coming back, how many mph do you average (not 50 mph).
In a situation such as this, average speed can be determined without
using distance. However, it’s better explained if distance is used.
Let distance going and returning, be D
Then time spent traveling outbound = 
Time spent traveling inbound = 
Total round-trip distance: 2D
Total time taken =  , or  , or  , or 
Therefore, average speed =  , or  , or  mph
You can do the check!!
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