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Compute the speed of the tip of the second hand

  1. Apr 13, 2005 #1
    I think this is a simple question but I'm not sure if I'm doing it right, this is what I have so far...

    A watch has a second hand 2.0 cm long.

    a) Compute the speed of the tip of the second hand.

    C = 2pi(r) r = 2.0cm = 0.02m - leave as cm or convert to m?

    V = [2pi(0.02m)]/60s = 2.1 x 10^-3 m/s - divide by 60s? :uhh:

    b) What is the velocity of the tip of the second hand at 0.0s ? At 15 seconds?

    well 0s=60s right...? if so then the answer is the same as a)

    for 15s V=[2pi(0.02m)]/15s = 8.3 x 10^-3 m/s

    c) Compute its change in velocity between 0.0 and 15 seconds.


    d) Compute its average vector acceleration between 0.0 and 15 seconds.

    I think I could answer this if I could figure out c)
  2. jcsd
  3. Apr 13, 2005 #2


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    Science Advisor

    Part A looks correct.

    For part B, remeber that velocity is a vector...amplitude and direction. Therefore the velocity at 0/60 sec is (using the i designator as a unit vector in the +x direction) [tex]\vec{v} = 2.1 x 10^{-3} \hat{i}[/tex]. At 15 seconds, the direction is going to be down (using j as the unit vector in the +y direction), in the -j direction with the same magnitude.

    Does that point out what you need to do for the rest of the problems?
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