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Kinematics-Boulder Falling Problem

  1. Jun 18, 2014 #1
    1. The problem statement, all variables and given/known data
    I'm not sure if this is the right place to post this but I need help understanding this question...
    Discounting air friction, approximately how far will the boulder have fallen in 2 seconds?

    a. 20m
    b. 45m
    c. 30m
    d. 90m

    2. Relevant equations

    x-x_0 = v_0t + 1/2at^2

    3. The attempt at a solution
    Now I know this is just a plug and chug equation and the answer is 45m (B). My question is more in understanding the physics behind using this equation....
    Originally i used a=Δv/Δt then plugged my found velocity into v=d/t but that gives me 40m, which is not an option.
    How do I know when to use which?

    Thanks for any help in clarifying this for me! :D
     
  2. jcsd
  3. Jun 18, 2014 #2
    "Discounting air friction, approximately how far will the boulder have fallen in 2 seconds?"

    Is that the entire question, or did they give any specifics as the the boulders initial speed? If the initial speed is zero, then you are considering tthe case in which an object has started at some position above the ground. You should be thinking about just as the boulder starts to fall, and what is happening during the fall.

    While the boulder is in free fall, what is governing the speed at which its falling?
     
    Last edited: Jun 18, 2014
  4. Jun 18, 2014 #3
    No other information was given. A picture of a boulder falling near a cliff was the only other thing given.

    But while the boulder is in free fall isn't it just gravity that's controlling the speed?
     
  5. Jun 18, 2014 #4
    Yes.


    what else do you know about the problem?

    EDIT: I'm sorry, what your first attempt should have given you is the correct answer as well. you may have just made an arithmetic error. both [itex] x = v_0t + \frac{1}{2}at^2 [/itex] and manipulating [itex] a = \frac{v_f - v_i}{t_f - t_i} [/itex] and using [itex] v = \frac{d}{t} [/itex] yield the same results
     
    Last edited: Jun 18, 2014
  6. Jun 18, 2014 #5
    Oh I see...well if the only force really acting on the boulder is gravity that means i have constant acceleration, right? And I'm now remembering that the correct equation for this problem is one used under constant acceleration...
     
  7. Jun 18, 2014 #6
    Yes, now if we let the downward direction be the positive direction, then we see that, like you said, [itex]a = g[/itex].

    so the equation now becomes [itex] x-x_0 = v_0t + \frac{1}{2}gt^2 [/itex].

    so you know that the only force acting on the boulder is gravity and thus it is experiencing constant acceleration. but you also know the initial speed, right?
     
  8. Jun 23, 2014 #7
    As he said earlier,the value of initial speed is not given. Only a diagram is given. So you can safely assume the velocity to be zero. If my assumption IS correct,then the answer is A) 20m/s
     
  9. Jun 23, 2014 #8
    Yes, That's what I was getting at and why I said SChiO must have made an arithmetic error. I just didn't want to give it away. I can only assume SChiO figured it out by now..
     
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