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Average velocity of bicycle

  1. May 30, 2009 #1
    1. The problem statement, all variables and given/known data
    A bicycle travels 3.2 km due east in 0.10 h, then 4.7 km at 15.0° east of north in 0.14 h, and finally another 3.2 km due east in 0.10 h to reach its destination. The time lost in turning is negligible. What is the average velocity for the entire trip?
    I need to find the magnitude and direction in degrees


    2. Relevant equations

    average velocity= rf-ri/t1-t2 but there is 3 different times and distances not two...

    3. The attempt at a solution

    I tried to do vector addition A+B and B+C then add them together but that didn't work to well.
     
  2. jcsd
  3. May 30, 2009 #2

    LowlyPion

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    Homework Helper

    You need to resolve the vectors into their x,y components, then determine the total displacement. That will yield your direction.

    For average V = |D|/Total time.

    Where |D| is the magnitude of your Displacement determined above, where

    D = A + B + C
     
  4. May 31, 2009 #3
    I found the initial velocity for x and y by using the formulas 4.7cos(15.0)= 4.539
    4.7sin(15.0)= 1.216
    delta t=0.10+0.14
    Then I found the x and y components by doing: delta rx= vix(delta t)+0= 1.089
    delta ry= viy (delta t)-1/2g(delta t)^2=-1.136
    Then I added A+B+C for x and y: x=3.2+1.089+3.2=7.48
    y=0+.0096+0
    Then added them and took the square root. Then divided that number by the total time. Didn't workout to well...
     
  5. May 31, 2009 #4

    LowlyPion

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    Ok. Try this way.

    A = 3.2 i + 0 j

    B = 4.7*Sin15 i + 4.7*Cos15 j

    C = 3.2 i + 0 j

    Then add them together

    D = (6.4 + 4.7*sin15) i + 4.7cos 15 j

    Your Δt is (.1 + .14 + .1)

    Your answer will be in km/h
     
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