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1D Motion

  1. Sep 23, 2010 #1
    1. The problem statement, all variables and given/known data
    Just when a Frisbee passes over the head of a student at a speed of 4m/s and decelerating at a rate of 1.5 m/s2, she starts to run accelerating at a rate of 1 m/s2. [neglect vertical motion of Frisbee]

    a)How long will it take the student to catch the Frisbee?

    b) If instead of accelerating the student were able to run at a constant speed of 2 m/s as soon as the Frisbee passes her head how far would she have to run to catch it?

    2. Relevant equations
    x=x0+V0t+1/2at2

    3. The attempt at a solution

    So for part a, I figured if I made both xfrisbee and xstudent equal to each other I could solve for t
    xfrisbee=0+4t+(0.5)(-1.5)t2
    xstudent=0+0t+(0.5)(1)t2

    0.5t2=4t+0.75t2
    t=0s t=3.2s

    And something similar for part b:
    xfrisbee=0+4t+(0.5)(-1.5)t2
    xstudent=0+2t+(0.5)(0)t2

    4t+(0.5)(-1.5)t2=2t
    t=0s t=2.7s
    x=(2m/s)(2.7s)=5.4m

    I just wanted to know if I got these right. Thanks!
     
  2. jcsd
  3. Sep 24, 2010 #2

    Doc Al

    User Avatar

    Staff: Mentor

    Looks perfect to me. :approve:

    (Nitpick: Careful not to round off until the last step--especially if you're submitting your work to an online system. They can be picky.)
     
  4. Sep 24, 2010 #3

    jhae2.718

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    Gold Member

    Looks like you did everything correctly.

    One caution: be careful with your signs; had you gone off of this equation, you would have gotten the result that the student will never reach the frisbee:
    What Doc Al says about rounding is good advice. The distance, if you carry figures through to the end, ends up at about 5.3333 m.
     
  5. Sep 24, 2010 #4
    Thank you for your help.

    I'll make sure to hold the rounding off until the end and to check my signs next time.
     
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