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Distance covered - kinematics

  1. Feb 3, 2015 #1
    1. The problem statement, all variables and given/known data

    A robotic bug is moving along the x-axis. Its equation of motion is x(t) = 10*t^2- 4*t - 6

    I have a problem with only one question:

    What is the distance covered by the bug between t = 0.1 s and t = 0.4 s?

    2. Relevant equations


    3. The attempt at a solution

    The first derivative of 10*t^2- 4*t - 6 is velocity: 20*t-4

    So I integrated 20*t-4 and computed the value of the integral from 0.1 to 0.4. I got 0.3 and it is wrong. I have no idea why. :(
     
  2. jcsd
  3. Feb 3, 2015 #2
    Why did you Integrated ??... I mean distance is asked and equation of x is given. So can't we just substitute the value and find the answer.
    Can you recheck the equation....
     
  4. Feb 3, 2015 #3
    I have found the value of displacement without a problem. But this is distance covered.
     
  5. Feb 3, 2015 #4

    haruspex

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    Right, but it's only in one dimension. What has to happen for distance covered and magnitude of displacement to be different?
     
  6. Feb 3, 2015 #5

    gneill

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    You differentiated and then integrated, which really just cancelled each other. But hold on to that derivative, it'll be useful :)

    The problem appears to be asking about the distance that the bug walked, not the displacement from its original location. That is, the length of the curve from where it started to where it finished. So, how do find the arc length of a curve?
     
  7. Feb 3, 2015 #6

    Orodruin

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    You have already gotten some sound advice. I am just going to offer an alternative way of thinking.

    If the displacement is not equal to the distance walked, then the bug must have turned around somewhere. How far did the bug have to walk to get there? How far did the bug have to walk back in order to make the displacement what it is?
     
  8. Feb 3, 2015 #7

    haruspex

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    Curve?
     
  9. Feb 3, 2015 #8

    gneill

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    In the general sense. A straight line is just a special case.
     
  10. Feb 3, 2015 #9
    The bug moves in X axis so it moves in straight line... Next (x) in distance or displacement... So the equation give in equation of distance. And the distance traveled by the bug is equal to displacement as the bug travels in straight line.
     
  11. Feb 3, 2015 #10

    gneill

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    If you walk straight across the room and back your displacement is zero (you're back where you started). Yet you have covered some distance while walking....

    So distance is not always the same as displacement.
     
  12. Feb 3, 2015 #11
    If there is no turning back then they are equal, right .....
     
  13. Feb 3, 2015 #12

    gneill

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    When motion is along a straight line that is true. So the question is, does the bug in the problem turn around?
     
  14. Feb 3, 2015 #13
    I don't know... the answer. But if the question doesn't say about turning back so can't be assume that it moves forward ???
     
  15. Feb 3, 2015 #14
    The bug must have turned back then. I got it. I have to digest it.
     
  16. Feb 3, 2015 #15
    What did you get, I am totally confused, Please explain ??
     
  17. Feb 3, 2015 #16
    I think I have figured it out.

    I have used the derivative to establish the turning point:

    20*t-4<0
    t<0.2

    I devided the path into two parts:
    (10*0.2^2-4*0.2-6)-(10*0.1^2-4*0.1-6) = -0.1

    (10*0.4^2-4*0.4-6)-(10*0.2^2-4*0.2-6) = 0.4


    So the distance covered would be 0.5.
     
  18. Feb 3, 2015 #17

    Orodruin

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    Yes, this looks fine.

    No, you have been told in the problem exactly how it moves by giving the position as a function of time. You can deduce from this whether it is turning around or not as the OP did.
     
  19. Feb 3, 2015 #18
    Many thanks. :) :) :)
     
  20. Feb 3, 2015 #19
    Thanks... It did raise my knowledge.... Thanks again to all
     
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