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Acceleration due to gravity/velocity questions (I've done the work)

  1. Feb 16, 2009 #1
    I'm having trouble with these 2 questions. I've tried them and come up with solutions but would like someone to check if I have done them properly.

    A traditional watch has a second hand 1.5 cm long from the center to the tip.

    a) What is the speed of the tip of the second hand?

    d = 2Pir
    d = 2(3.14)(1.5)
    v = d/t
    v = 2(3.14)(1.5)/60s
    v = 0.157 cm/s

    b) what is the velocity of the tip at 15s? 45s? 60?

    0.157 cm/s (right)
    0.157 cm/s (left)
    0.157 cm/s (up)

    c) What is the change in velocity between 30s and 45s?

    I'm having trouble with this one. Do I use the pythagorean theorem because one is pointing to the left and one is pointing down.

    0.157^2 + 0.157^2 - sq root
    v = 0.222 cm/s (sw)

    A student tosses their pen vertically upward with an intial velocity of 3.8 m/s.

    a) Maximum height

    Vf^2 = Vi^2 + 2ad
    0 = 3.8^2 + 2(-9.8)d
    d = 0.736735 m

    b) How much time will pass before the pen returns to his hand if it is at the same level that he released the pen at?

    d = v2t - 1/2at^2
    0= (3.8)(t) - 1/2(9.8)t^2
    t = 0.775510 s

    c) if the student's hand is 1.5 m above the ground and he misses the pen, with what speed will the pen hit the ground.

    0.736735 + 1.5 = 2.236735 m

    Vf^2 = Vi^2 + 2ad
    Vf^2 = (0) + 2(9.8)(2.236735)

    vf = 6.62117 m/s (down)

    Can someone plz check these over. I would really appreciate it. Thank you.
  2. jcsd
  3. Feb 16, 2009 #2


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    1b you have the positions not the velocity. When the hand is at 15s=3 O'clock, which direction is it moving in?
  4. Feb 16, 2009 #3

    Are the other questions correct?
  5. Feb 16, 2009 #4


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    1a is correct.
    1b draw a diagram and put an arrow for the direction the hand is moving - is it moving up/down/left/right?
    1c, correct but possibly by accident. Draw a diagram with the speed in x and y axis and the overall speed is the hypotenuse of the triangle

    2a correct
    2b correct you could also do v=u+at for the upward path and then double the time.
    2c you can also use the fact that it arrives back at the hand with the same speed it was thrown up with. So it's the same as throwing it down with an initial speed of 3.8m/s
  6. Feb 16, 2009 #5
    Thank you so much I really appreciate it :)
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