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Physics Finding Total Distance?

  1. Mar 26, 2012 #1
    1. The problem statement, all variables and given/known data
    John had 5 dogs pulling his sled at full speed for 24 hours. 2 dogs than escaped. however, the journey continued at 3/5 the original velocity and reached his destination 48 hours later. if the two dogs did not escape for an additional 50km he would have reached 24 hours late instead. What is the distance he covered??
    Please explain...




    3. The attempt at a solution
    Well so far heres what i got:
    I think i can assume(not sure) that with 5 dogs he traveled 50km in 24 hours according to the last bit of information.

    So than that means in the previous 24 hours he would have went 50km with 3 dogs - this is what i dont understand, how can 5 dogs travel 50km in 24 hours while 3 dogs travel 50km in 24 hours as well??

    And therefore in the first 24 hours the dogs traveled (50/3) x 5 = 83.333km

    and for my total I get 50 + 83.333 = 133.33

    Please explain to me the part where 3 dogs travel the same distance in the same time as 5 dogs...
    Thanks!

    The possible answers are 100, 133, 167, 200, or 267 (all kilometers)
     
    Last edited: Mar 26, 2012
  2. jcsd
  3. Mar 26, 2012 #2

    gneill

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    Staff: Mentor

    No, you don't know how long it took the five dogs to cover the 50km. All you know is that it takes a total of 24 hours to cover the "second stretch" if the first 50km are done with a full team of 5 dogs, and 48 hours if done with 3 dogs.

    So if v is the "top speed" with 5 dogs, then to cover the second stretch:
    $$\frac{50 km}{v} + \frac{x}{\frac{3}{5}v} = 24 hours$$
    where x is the currently unknown portion of the second stretch: d2 = 50 km + x

    You should be able to write another equation for d2 for the case where only three dogs are employed for the whole second stretch. Substitute for x and d2 accordingly, solve for v.
     
  4. Mar 26, 2012 #3
    Ok, so that means i would have 3 equations:

    Equation 1. 50km/v + x/3/5v = 24hours
    Equation 2. d2 = 50km + x
    Equation 3. d2(with 3 dogs) = 3/5v x 48hours

    Correct?
    If correct, than:
    Equation 3: d2 = (144/5)v
    And plug that into Equation 2 which is d2=50km + x
    So sub in (144/5)v for d2 and solve for x which means: x = (144/5)v - 50km
    And than sub the new equation 2 into equation 1 and solve for v?

    So.. 24= ( 144/5v - 50km)/(3/5)v + 50/v

    Which turned out to be 1.3888km/h

    So than in the first 24 hours he traveled 1.3888 x 24 = 33.333km?
    and than in the 48 hours with 3/5 dogs he traveled 1.38888x48 = 39.999
    so in total it equals 73.333km??

    I know im wrong but i dont know where? Please reply back..
     
  5. Mar 26, 2012 #4

    gneill

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    Well, the results look reasonable to me. Why do you think you're wrong? can you find a contradiction between the results and the given information?
     
  6. Mar 26, 2012 #5
    Its just that i have the possible answers and 73.333km isnt one of them, is there anywhere i could have gone wrong??
     
  7. Mar 26, 2012 #6

    gneill

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    I don't think so, your calculations looked okay to me (at first glance at least!). I'll give it some thought...
     
  8. Mar 26, 2012 #7

    gneill

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    Perhaps I've made a misinterpretation of the problem text? When you wrote:
    Did you mean "late" or "later"? I had assumed a typo, given the previous sentence :redface: In other words, would he have arrived in the next 24 hours, or would he arrive 24 hours AFTER he would have arrived with a full team (24 hours "late")?
     
  9. Mar 26, 2012 #8
    Ok thanks!, but can you tell me why i cannot assume the last part as the dogs travelling 50km in 24 hours?
     
  10. Mar 26, 2012 #9
    No its meant to be he arrived only 24 hours late rather than the 48 hours he would have been late

    So if he had 5 dogs for an additional 50km he would have been ONLY 50km late

    Hope that clears up some misinterpretations
     
  11. Mar 26, 2012 #10

    gneill

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    Because it wasn't stated as such :smile: All you know is that at one speed it takes a given time, and doing a portion at another speed it takes a different time. You can't assume that the distance was half just because the time was halved.
     
  12. Mar 26, 2012 #11

    gneill

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    Really, that only helps a bit :smile: "50km late"?
     
  13. Mar 26, 2012 #12
    Oops sorry i meant ONLY 24 hours late... Sorry...
     
  14. Mar 26, 2012 #13

    gneill

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    So, just to be clear, in the first case he is 48 hours late, and in the second case he would be only 24 hours late. This would be based on the expected arrival time of T = D/v where D is the total distance and v the velocity with a full team for the entire trip. Is that the scenario?
     
  15. Mar 26, 2012 #14
    Ya thats what i believe the scenario is.
     
  16. Mar 26, 2012 #15

    gneill

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    Okay, then write equations for elapsed times. Let T be the expected time, v be the normal velocity, d1 the first distance using all the dogs, d2 the remaining distance. D is the total distance.

    Normal travel would yield an "expected" elapsed time of T = D/v. How would you write equations for the described cases? One elapsed time is T + 48hr, and the other is T + 24hr. What distance and velocity expressions would correspond to them? Hint: dealing with the 50km is the tricky bit :wink:
     
  17. Mar 26, 2012 #16
    sooo.. not so sure about this but: d1 = v(T+24)? and d2 = 3/5v(T+48)?
    and honestly i have no idea for the 50km but is it d1 + 50km =T??
     
  18. Mar 26, 2012 #17

    PeterO

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    A picture/graph might help.

    With all 5 dogs, you would have a flat v-t graph for the entire trip - ending at some time t.

    For the trip with the 2 dogs running away, it would be flat at the original speed for 24 hours, then drop to 3/5 ths of that speed for extra time - finishing at t + 48

    Draw those two graphs on the same axes, and you see given a common trip, the area under each graph has to be the same.
    The extra 48 hours at 60% of original speed, makes up for the missing section from 24 hours to t - which amounts to 40% [the top bit] of the distance covered if all dogs had stayed.

    Now draw in the third possibilty - with the 5 dogs staying for an extra 50km. But that extra 50 k is done at full speed.

    The distance covered in 24 hrs = only 40% of 50km.

    The
     
    Last edited: Mar 26, 2012
  19. Mar 26, 2012 #18

    gneill

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    No, d1 is the distance covered in the first 24 hours. So d1 = v*24hr. You could also say that the time to cover distance d1 is d1/v = 24hr.
    You can't equate distance with time...the units don't match.

    If the two sections of the trip described in the first scenario are d1 and d2, the first traveled at speed v and the second at (3/5)v, what are the elapsed times for each (symbolically)?

    If d1 is made 50km longer, then d2 must be made 50km shorter. What are the new elapsed times for each (symbolically)?
     
  20. Mar 26, 2012 #19
    So, would the graph look like the picture i included? and Are my equ Untitled-1.png ations correct?
     
  21. Mar 26, 2012 #20

    PeterO

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    In case you are not familiar with solving this type of problem with v-t graphs, the attached file shows you them.
    I find them particularly useful.
     

    Attached Files:

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