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Changes in velocity with Direction change

  1. Mar 6, 2013 #1
    1. The problem statement, all variables and given/known data

    Tim is running cross country at 6.4m/s when he completes a wide angle turn and continues at 5.8m/s[w]. What is his change in velocity?


    2. Relevant equations

    Δv=v2-v1
    a2+b2=c2

    3. The attempt at a solution

    i am not sure how i am supposed to find the change in velocity...
    i could just find the hypotenuse but i think that is giving me the resulting acceleration and not the actual change. but im quite sure its not as simple as just subtracting the two values either ??
     
  2. jcsd
  3. Mar 6, 2013 #2
    You have to remember that velocity is a vector: so we have [itex]v_1 = (0,6.4)[/itex] and [itex]v_2 = (5.8,0)[/itex] (where [itex](1,0)[/itex] points West and [itex](0,1)[/itex] points South), so as you correctly wrote: [itex]\Delta v = v_2 - v_1 = (5.8, -6.4)[/itex]. This is technically the change in his velocity, but the question may be just asking for its magnitude (which, as you note, is the hypotenuse).
     
    Last edited: Mar 6, 2013
  4. Mar 6, 2013 #3
    ok i will include both in my answer then. thank you. also should i be including the angle/direction? do you draw the s or w arrow first when drawing the diagram? like will the angle be between the hyp and 5.8 or the hyp and 6.4?
     
  5. Mar 6, 2013 #4
    If you include the information about what your axes are (i.e. South = (0,1), West = (1,0) ) then when you write the components of [itex]\Delta v[/itex] down you have represented it unambiguously. You shouldn't need to include it unless the question specifically asks you for it.

    I'm not sure I understand the first question. In answer to the second, you would have to be specific (e.g. "the vector is 10° to the East of North", or "at a bearing of 010°" or some similar specification of angle - and note that was just an example, not the direction of [itex]\Delta v[/itex])
     
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