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Average speed? -- Homework problem

  1. Jan 21, 2015 #1
    Ok let me just type the question and get to the real question I have.
    If you traveled one mile at a speed of 100 miles per hour and another mile at a speed of 1 mile per hour, your average speed would not be (100 mph+ 1 mph)/2 or 50.5 mph. What would be your average speed? (Hint:what is the total distance and total time)
    Ok so I worked it out, solved for the individual times and got 1/100 hr for one time and


    1 hr for the second time. Then I added those times and divided by the total distance , 2 miles. And I got around 1.98 mph as the average speed



    My question. Why CANT do the whole (100+1)/2? Or rather why don't you get the correct answer? Doesn't it make sense if they're asking you for your average speed to average out the two average speeds they have you?
    And also, one of the speeds is 100 while the other is 1... So it makes no sense to me that the "final " average speed be 1.98? Can someone make sense of this for me? It just seems illogical .. Or maybe since it's just my first week in pre ap physics I'm just not seeing something critical ...?
     
  2. jcsd
  3. Jan 21, 2015 #2
    Sorry for the grammar/spelling mistakes . I'm using my iphone(autocorrect )
     
  4. Jan 21, 2015 #3

    Bystander

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    Speed is defined as the magnitude of the change in displacement of position (distance) divided by time. "s = d/t"
    If you want to know distance or time, you rearrange to "t = d/s" or "d = st."
    Use your method for average speed and see if you can recover distance or time for the two mile journey.
     
  5. Jan 21, 2015 #4
    But the thing I'm trying to understand is how is it possible that your average speed is 1.98 when for one mile your speed was 100 miles per hour and for the next mile it was 1mph ....how does that logically make sense?? How can your AVERAGE speed be barely 2?
     
  6. Jan 21, 2015 #5

    Bystander

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    How did the "tortoise beat the hare?" Do the 100 mph for the first 10 feet of the two mile journey. Your method for average speed hasn't changed. The total trip time has nearly doubled. The actual average speed has been cut in half. You're not interested in averaging instantaneous rates. If you travel one foot at 100 mph, and the second foot, and the third, and the fourth, you can "prove" that you averaged 100 mph for the whole trip rather than 50-whatever. You are after a "weighted" average of rate for each leg times the elapsed time for each leg or for total distance travelled.
     
  7. Jan 21, 2015 #6
    Hmm...I read your post and it made sense for just a second lol..but then whatever logic burst into my head burst right out. The 100 and the 1 equaling the 2 is what's confusing me. Could you maybe explain it in the context of the question? (Not if it's too much)

    edit: I just noticed you pretty much did answer in the context of the question, nevermind, sorry haha.
     
    Last edited: Jan 21, 2015
  8. Jan 21, 2015 #7

    BvU

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    Dear Rasofia, welcome to PF, by the way. :)

    I recognize your burden: When on holiday in France, I cycle uphill for an hour at 5 km/h (barely enough not to keel over), then I race down at a totally irresponsible 60 km/h (fortunately the road is wide). Takes me an hour the get up the 5 km long bloody slope and just five minutes to get down. 65 minutes for 10 km. A disappointing average of 9.2 km/h. And I realize that if I would have done it in zero time, my average would still have been no more than 10 km/h.

    My kid brother lives in the Netherlands. He has no problem whatsoever with this: He goes out training for two hours. First hour he's racing like an idiot at 40 km/h. Second hour he takes it easy and dreams of a cool beer at 20 km/h. By the time he gets home he calculates 1 hour at 40 km/h and one hour at 20 km/h, so that's an average of (40+20)/2 = 30 km/h.

    How come that to find an average speed he can do what you and I would like to do, but apparently can't do ?

    Time for some math. But not now (it's late in this time zone). Perhaps you can set up the (clearly different) calculations for the averages in these two long-winded cases ;) ( I wanted a sleepy smiley, but can't find one)
     
  9. Jan 21, 2015 #8
    Thank you for the time and response! And yes, that's what I would've thought if I were your brother. To just add them and divide by two, because that to me would make sense...but now I see, really, that's not the case
     
  10. Jan 21, 2015 #9

    Bystander

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    BvU got you where you need to be with the idea?
     
  11. Jan 21, 2015 #10
    It's still a bit baffling but I'll handle, haha. Processing takes a bit of time. Thanks to you both!
     
  12. Jan 21, 2015 #11

    BvU

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    It's really math, you know: In the second case you have $$v_{avg} = distance_{tot}/time_{tot} = {v_1*time_1 + v_2*time_2\over time_1+time_2}$$Because the times are equal, this simplifies to $$v_{avg} = {v_1 + v_2\over 2}$$
    And guess what: in the first case you have $$v_{avg} = distance_{tot}/time_{tot} $$ as well. But now writing that out gives $$v_{avg} = {distance_1 + distance_2 \over distance_1 /v_1 + distance_2 /v_2}$$This time, the distances are the same, so it simplifies to$$v_{avg} = {2 \over 1 /v_1 + 1 /v_2}$$which looks ugly.

    Another way to present that is$$ {2 \over v_{avg} }= {1 \over v_1} + {1 \over v_2}$$

    So you see this factor of two appear as a limit when ##v_2 \rightarrow \infty##. No point in risking 70 km/h downhill just to improve your average speed :(

    :)
     
  13. Jan 21, 2015 #12

    haruspex

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    I gather that elite cyclists try to maintain a constant speed relative to the wind in time trials. Even I try to put the effort into the uphill (where there's usually some shelter from any headwind) but tuck in and coast downhill.
    But hey, cycling in France? Why go fast at all? Enjoy the scenery.
    ##{2 \over v_{avg} }= {1 \over v_1} + {1 \over v_2}##​
    I'm sure you know, but Rasofia might not, that this is the harmonic mean. Note the analogy with parallel resistors.
     
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