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Average speeds of back and forth trip 
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#1
Aug1111, 05:09 PM

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1. The problem statement, all variables and given/known data
You drive on Interstate 10 from San Antonio to Houston. Half the time at 55 km/h, and the other half at 90 km/h. On the way back, you travel half the distance at 55 km/h and the other half at 90km/h. What is the average speed of the trip from San Antonio to Houston? What is the average speed of the return trip from Houston to San Antonio? What is the average speed of the whole trip? 3. The attempt at a solution From San Antonio to Houston, half the time is traveled between the two speeds. So: d / (1/2)(d/55) + (1/2)(d/90) 68.27586 km/h average speed. From Houston to San Antonio, half the distance is traveled at each speed. ((1/2)55t + (1/2)90t) / t =(1/2)55 + (1/2)90 = 72.5 average speed. My book seems to have the two answers reversed. Did I do these correctly? 


#2
Aug1111, 05:26 PM

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I added the extra  necessary  parentheses in the following. d / ((1/2)(d/55) + (1/2)(d/90)) What is (d/2)/55 ? ... It's the time required to travel a distance d/2 at a speed of 55... etc ... d/(total time) is average speed. So this is not the case of: "half the time is traveled between the two speeds". 


#3
Aug1111, 05:48 PM

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I am not sure what you are saying..
Does my work not agree with total distance / total time? Why are you telling me that? Also, why isn't it a case of traveling half the time at each speed? That's explicitly what the problem says. 


#4
Aug1111, 06:10 PM

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Average speeds of back and forth trip
So you are finding the time for half the distance at 55 + half the distance at 90. 


#5
Aug1111, 06:21 PM

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Now I'm really confused. If time is (d/r), (1/2)(d/r) or (d/2r) is half of the time is it not??
Wait, I think I'm starting to get it. d/r is the time spent traveling d at r, so half of it is not half of the total time, only half of the time spent traveling d at r. Thanks! 


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