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Net displacement problem with two triangles

  1. May 27, 2013 #1
    1. The problem statement, all variables and given/known data

    A ball on a porch rolls 60 cm to the porch's edge, drops 40 cm, continues rolling on the grass, and eventually stops 80 cm from the porch's edge.

    What is the magnitude of the ball's net displacement, in centimeters?



    ball starts rolling here
    \/
    rolls 60cm->
    #----------#
    ******** -
    ******** - <- ball goes down 40 cm
    ******** -
    ******** #-----------------# <---ball stops rolling here

    ..................ball rolls 80cm ^


    2. Relevant equations

    The length of both a and b are known for both triangles, then c, the hypotenuse, can
    ...................................... _________
    be calculated as follows: c=√a^2+b^2.


    3. The attempt at a solution

    72 cm and 89 cm are the hypotenuse's of two different triangles.

    The first triangle is 40 and 60; The hypotenuse is 72 cm. The second is 40 and 80: this one's hypotenuse is 89 cm. I need to find the magnitude of the ball's net displacement. This is where I'm confused. And have I done this problem correct up to this point or am I way off?
     
    Last edited: May 27, 2013
  2. jcsd
  3. May 27, 2013 #2
    Why do you care about a point in the middle?
     
  4. May 27, 2013 #3
    I don't know. I may be over thinking this problem
     
  5. May 27, 2013 #4
    For displacement, only the end points are important. Consider them.
     
  6. May 27, 2013 #5
    Well I would say that where the ball started is the first point and the end point is where the ball stopped on the lawn. And so the displacement would probably be a straight line from the start to finish.
     
    Last edited: May 27, 2013
  7. May 27, 2013 #6
    That is correct.
     
  8. May 27, 2013 #7
    Ok. I understand the concept then. I just don't understand how to get the solution.
     
  9. May 27, 2013 #8
    Why exactly can you not compute the displacement? You were able to compute displacements to/from some arbitrary middle points.
     
  10. May 27, 2013 #9
    I'm not sure.
    Is it the displacement of the first triangle(72cm) + 80cm for the remaining length traveled on the x axis?

    152 cm?
     
    Last edited: May 27, 2013
  11. May 27, 2013 #10
    Again. You can compute the displacement to some arbitrary middle point ("the first triangle"). What can't you just do it for the entire displacement in the exact same way?
     
  12. May 27, 2013 #11
    vector addition?
     
    Last edited: May 27, 2013
  13. May 27, 2013 #12
    I guess I don't know the correct formula.
    Am I correct in thinking that the unknown displacement wouldn't be part of a right triangle, and so I can't use the Pythagorean theorem?
     
  14. May 27, 2013 #13
    Look at the end points of the displacement in the diagram in your post #1. Then look at the point at the intersection of the vertical line from the first point, and the horizontal line from the second point. What can you say about the triangle made of these three points?
     
  15. May 27, 2013 #14
    Oh okay thankyou. So we can add 60 +80. and thats 140. and the y axis is 40. so the displacement is 145.6cm.

    it has 2 sig figs so would it be: ≈145 or ≈146 or ≈145.60 ?
     
  16. May 27, 2013 #15
    Very well. You should round it to the nearest integer, because your input did not have any meaningful decimal figures. You have .6 - does that round toward 0, or toward 1?
     
  17. May 27, 2013 #16
    toward 1
     
    Last edited: May 27, 2013
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