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    ConfusedAutumn's Avatar
    ConfusedAutumn Posts: 4, Reputation: 1
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    #1

    Jan 28, 2010, 01:00 PM
    How to calculate average speed
    A cat travels across gardens at 240 metres an hour and returns on the same route at 160 metres per hour -- what is his average speed?
    ebaines's Avatar
    ebaines Posts: 12,131, Reputation: 1307
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    #2

    Jan 28, 2010, 01:30 PM

    Remember the basic equation of velocity:



    Here we want to find the value for velocity, so rearrange:



    So we need now to find the distnace traveled, and the time it takes. Let's call the width of the garden W. Then the total distance traveled is 2*W, so that's the numerator. Now you need to figure out how long the total trip took to complete - which is the sum of the time for the trip over plus the trip back. The time for each segment of the cat's journey is found using the formula time = distance/velocity; do you see how I got that? So the trip over takes W/240, and the trip back took W/160, and the total time is W/240 + W/160. Now you can find the average speed:



    See how the W terms cancel out? Can you take it from here, and post back the final answer that you get?
    ConfusedAutumn's Avatar
    ConfusedAutumn Posts: 4, Reputation: 1
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    #3

    Jan 29, 2010, 09:00 AM
    Quote Originally Posted by ebaines View Post
    Remember the basic equation of velocity:



    Here we want to find the value for velocity, so rearrange:



    So we need now to find the distnace traveled, and the time it takes. Let's call the width of the garden W. Then the total distance traveled is 2*W, so that's the numerator. Now you need to figure out how long the total trip took to complete - which is the sum of the time for the trip over plus the trip back. The time for each segment of the cat's journey is found using the formula time = distance/velocity; do you see how I got that? So the trip over takes W/240, and the trip back took W/160, and the total time is W/240 + W/160. Now you can find the average speed:



    See how the W terms cancel out? Can you take it from here, and post back the final answer that you get?
    The average speed is 200 meters I think?
    Unknown008's Avatar
    Unknown008 Posts: 8,076, Reputation: 723
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    #4

    Jan 29, 2010, 09:29 AM

    Nope, try again.

    Simplify the fraction just below there;

    ebaines's Avatar
    ebaines Posts: 12,131, Reputation: 1307
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    #5

    Jan 29, 2010, 09:32 AM
    Quote Originally Posted by ConfusedAutumn View Post
    The average speed is 200 meters I think?
    Sorry - that's incorect. You can't just average the two speeds. You can see why that doesn't work if you consider an easier case. Suppose you go 60 miles at 60 MPH, taking 1 hour of time. Then you cut your speed to 1 MPH and go another 60 miles - that will take another 60 hours, right? Total elapsed time is 1 hour + 60 hour= 61 hours, and total distance traveled is 2*60 = 120 miles. So your average speed is 120 miles/61 hours = 1.97 MPH. Compare that to simply averaging the speeds: (60+1)/2 = 30.5 MPH. Clearly averaging the speeds gets you the wrong answer.
    ConfusedAutumn's Avatar
    ConfusedAutumn Posts: 4, Reputation: 1
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    #6

    Jan 29, 2010, 11:36 AM
    Quote Originally Posted by ebaines View Post
    Sorry - that's incorect. You can't just average the two speeds. You can see why that doesn't work if you consider an easier case. Suppose you go 60 miles at 60 MPH, taking 1 hour of time. Then you cut your speed to 1 MPH and go another 60 miles - that will take another 60 hours, right? Total elapsed time is 1 hour + 60 hour= 61 hours, and total distance traveled is 2*60 = 120 miles. So your average speed is 120 miles/61 hours = 1.97 MPH. Compare that to simply averaging the speeds: (60+1)/2 = 30.5 MPH. Clearly averaging the speeds gets you the wrong answer.
    Ok thank you, I think I understand it now, is the answer 192?
    Unknown008's Avatar
    Unknown008 Posts: 8,076, Reputation: 723
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    #7

    Jan 29, 2010, 11:42 AM

    Ah! That's it! Well done! :)
    ConfusedAutumn's Avatar
    ConfusedAutumn Posts: 4, Reputation: 1
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    #8

    Jan 29, 2010, 12:06 PM
    Quote Originally Posted by Unknown008 View Post
    Ah! That's it! Well done! :)
    Thank you for all your help :D xx

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