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Problems with velocity/acceleration involving clock arm

  1. Oct 4, 2011 #1
    1. The problem statement, all variables and given/known data
    A traditional watch has a second hand 2.0 cm long, from centre to tip

    a. what is the speed of the tip of the second hand?
    b. What is the velocity of the tip at 20s? 40s? and 60s?
    c. What is it's change in velocity between 15s and 30s?
    d. What is it's average acceleration durning the same time interval


    2. Relevant equations
    d = v * t
    sine law
    cosine law
    SOH CAH TOA

    3. The attempt at a solution
    a) used SOH CAH TOA to find the third side of the 90 degree (clock) triangle
    cos45 = 2/h
    cos45h = 2
    h = 4cm (rounded)

    v = d/t
    v = 4cm / 15s
    v = 0.3cm/s

    therefore the velocity of the clock arm is 0.3cm/s
    ****attached is the clock and the final 90 degree triangle****

    b) I used the cosine law to TRY to find the other side
    due to the clock being on 20s the angle from the start to the ending point of 20s is 120 degrees

    C^2 = A^2 + b^2 - 2(2)(2)cos120
    c^2 = 8 - 4
    C^2 = 4
    C = 2

    this is where I got stuck
    ****there is an attachment of the triangle I drew****

    thanks to anyone willing to help :D
     

    Attached Files:

  2. jcsd
  3. Oct 4, 2011 #2
    What is the difference between speed and velocity?
     
  4. Oct 4, 2011 #3
    the only difference I really see would be that a velocity is a vector while speed is a scalar.
     
  5. Oct 4, 2011 #4

    SammyS

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    What is important is the actual distance that the tip of the second hand travels, not the distance between one location and another (which is displacement, not distance traveled).
     
  6. Oct 5, 2011 #5
    I'm still pretty confused on how to do it :/
     
  7. Oct 6, 2011 #6
    bump
    (come on I need lots of help!)
     
  8. Oct 6, 2011 #7

    SammyS

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    How far does the tip of the second hand travel in 60 seconds?
     
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