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Homework Help: Physics homework help? Grand Canyon problem

  1. Feb 4, 2015 #1
    1. The problem statement, all variables and given/known data
    If you were to throw a large log over the edge of the Grand Canyon and it took 5.65 seconds to hit the ground, calculate the velocity of the log at impact in m/s and calculate the distance the log fell in feet.

    2. Relevant equations
    vf=vi + at, s=d/t


    3. The attempt at a solution
    So, I don't see any initial velocity in the question, which is weird because it says 'throw'..so that's a bit confusing. Do I just make it zero? Well, if I did, I'd get vf=55.4m/s. Now, for the distance part, I'm assuming I could just use the speed formula? (55.4)=d/(5.65) , and then I get 313.01 meters...(and an even larger number when converted into feet)...the problem is that the answer is 156.4 m.....how????
     
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  3. Feb 4, 2015 #2

    Dick

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    The time is good. That distance formula is only used is the speed is a constant. You are dealing with uniformly accelerated motion. Can you find another formula for that?
     
  4. Feb 4, 2015 #3

    v=deltax/t ? maybe deltax=.5(vi+vf)t ?
    I don't even know what I'm doing, to be honest.
     
  5. Feb 4, 2015 #4

    SammyS

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    Presumably, you "throw" the large log horizontally, so the vertical component of velocity is zero initially.

    The final velocity (vertical component) may be 55.4 m/s, but that's not the velocity during the entire 5.65 seconds.

    What's the average velocity if the acceleration is constant?
     
  6. Feb 4, 2015 #5

    so why doesn't v= deltax/t work? or rather, s=d/t ?
     
  7. Feb 4, 2015 #6

    Oh wait a second, s=d/t and v=deltax/t ...those are equations for average speed and average velocity?? I always assumed they could be for either instantaneous velocity or speed...my teacher never clarified, he just gave us equations.
     
  8. Feb 4, 2015 #7

    gneill

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    The formula v = d/t only applies to un-accelerated motion. But
    s = d/t applies when the velocity is constant. Here the velocity is not constant: the log accelerates due to gravity as it falls. You need another Relevant equation, one that predicts the distance traveled when the object is undergoing acceleration. Check your text or class notes.
     
  9. Feb 4, 2015 #8
    yes, I ended up using deltax=.5(v1+vf)t ...The only thing is that displacement (deltax) and distance are NOT the same things, right? So, how is it that I can use this formula and get the right answer?
     
  10. Feb 4, 2015 #9
    [Moderator note: The appearance of another template here is not due to a new problem being posted, it's the result of a thread merge]


    1. The problem statement, all variables and given/known data

    If you were to throw a large log over the edge of the Grand Canyon and it took 5.6 seconds to hit the ground, calculate the velocity of the log at impact in m/s and calculate the distance the log fell in feet.

    2. Relevant equations
    s=d/t v=deltax/t deltax=.5(vi+vf)t vf=vi+at



    3. The attempt at a solution
    Yes, I know, it's a simple question, but I'm just CONFUSED. Simply confused at everything in physics right now and I have a test tomorrow. Here are some questions I'm having just for this problem:
    First off, it says 'throw', but there's no initial velocity mentioned, so I'm assuming I just label vi as zero?
    Second, it would be vf=at, therefore (9.8)(5.65) ...would 9.8 be negative?? If so, why?
    Third, once you find the velocity 'at impact' (which is 55.4), why is it that you can't just plug it into s=d/t and find distance? Or into v=deltax/t? I just realized, fyi, that the formula s=d/t refers to AVERAGE speed, not instantaneous speed as well right? Same with average velocity. My teacher simply gave us the equations and said to plug in givens. Well anyways, is that why we don't use those formulas, because they are AVERAGES? And we only have a velocity/speed for a SPECIFIC time?
    Third, because of what I just mentioned, then, I'd use a different formula, maybe deltax=.5(vi+vf)t ...but the thing is, it asks for DISTANCE. And displacement(deltax) and distance are different, are they not??

    Thanks for your time and sorry...I'm just SO confused right now and even more anxious...I have a test tomorrow! :(


    I guess my MAIN question here is why is it that you can use a displacement formula when you're looking for distance?
     
    Last edited by a moderator: Feb 5, 2015
  11. Feb 4, 2015 #10

    phinds

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    Plugging things into formulas is a waste of time if you don't understand what the formulas MEAN. That's what you should be studying if you want to learn anything. See if somewhere in your text it doesn't have some explanation of how the various formulas come to be.
     
  12. Feb 4, 2015 #11

    The thing is I have no book and my notes are just FULL of formulas and examples, that's all. That's why I'm on this website trying to ask around and understand what this all means and what exactly I'm even doing.
     
  13. Feb 4, 2015 #12

    SteamKing

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    Hint: Here, 'distance' = 'displacement'.
     
  14. Feb 4, 2015 #13

    phinds

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    That is most unfortunate. I do not envy you your task of learning anything under those circumstances. What kind of course are you taking that you do not have a text?
     
  15. Feb 4, 2015 #14
    Well, how do I know that it's implied that they equal on certain problems? For example, if I asked you what the distance and displacement were after running 2 laps (and say each lap is a quarter of a mile). Well, the distance would be 1/2 mile..BUT the displacement would be zero since your returned to your starting point.
    There, distance and displacement certainly aren't the same.
     
  16. Feb 4, 2015 #15
    Pre-ap physics. My teacher feels it's better not to go by the book so we never checked out any.
     
  17. Feb 4, 2015 #16

    gneill

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    In this case that formula would work because the initial speed is zero, so the average speed is one half the final speed.

    It just so happens that the area under a velocity vs time graph gives the distance traveled. For a constant acceleration the plot would be a straight line (v = a*t), so the area under that line would be the area of a triangle with base t and height vf. That is, d = 0.5*vf*t.
     
  18. Feb 5, 2015 #17

    haruspex

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    It's (partly) the difference between distance travelled in whatever directions and distance in a specific direction. It says 'distance the log fell'. OK, you could argue that as 'distance travelled while falling', but I think a more natural interpretation is distance of movement in the vertical direction.
    Since it seems to have been thrown horizontally (or there's not enough information to solve it), distance travelled vertically will equal vertical displacement - there's no backtracking in the vertical direction.
     
  19. Feb 5, 2015 #18

    haruspex

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    Last edited by a moderator: Feb 5, 2015
  20. Feb 5, 2015 #19

    SteamKing

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    In the context of the OP, the log was not running around a track.

    That's why I said, 'Here, 'displacement' = 'distance'.

    If you come to PF looking for help, it's not a good idea to be snarky with people who volunteer their time to help others. ;)
     
  21. Feb 5, 2015 #20
    I wasn't being snarky. I'm actually very appreciative of everyone that's taking their time to comment, including you because you actually just opened my eyes to something that I couldn't see. So thank you very much for that and I apologize if I came off as rude. (That's why I don't like to text, people have told me I come off that way).
     
  22. Feb 5, 2015 #21

    BvU

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    Dear Rasofia,

    I sympathize. Hope you can come to rest a little bit and be fresh for the test tomorrow. In your circumstances, that seems to be the most important thing you can influence.

    PF only hears -- reads -- one side of the whole happenstance. But if you have no textbook or authorized lecture notes, life can become pretty difficult once you start lagging because it takes a while to understand the underlying concepts.

    As you already discovered, messaging on a forum is not the ideal way to bring across these concepts -- too easy to get distracted by misunderstanding one another and not realizing it. When I read through your threads, I estimate you have had really excellent support, it's just not quite what you needed most: direct contact to ensure the concepts come across and are understood thoroughly.

    I also want to bring forward a few items in defence of teacher -- but I may be wrong because I don't have much to go on. Often teachers who work without textbooks are the most dedicated and committed ones. But they do depend on students actively participating. If that's the case for you, all I can advise is to become involved more intensely in class, and ask for clarification if the explanations are unsatisfactory. Not much use now, but perhaps in the future -- it's only the beginning of February now.


    Oh, and in your case a lot of your problems originate in the confusion between vectors and scalars. Vectors have a magnitude and a direction. Scalars are numbers, generally with a dimension. The magnitude of a vector is a scalar. Vectors mostly live in a coordinate system with agreed upon orientation (e.g. up = positive). A component of a vector is a scalar too. Magnitude can not be negative, but a component can. It's all basic math; physics is built upon such foundations.

    Language often falls short of conveying the exactitude of math and physics. Example: your exercise is not complete, but depends on (reasonable) assumptions, such as the departure of the log with zero vertical speed (vertical speed is a scalar -- that's why I use the word speed). In the text the word heavy indicates that. But not neccessarily with zero velocity (velocity is a vector -- in this case you want to throw the log away in a horizontal direction so it doesn't roll down, but experiences free fall under gravity. After all, the canyon walls are not vertical!): however, the horizontal speed ( :) ) has no influence on the time needed to reach the bottom.

    But then again, where the exercise should ask for speed hitting the ground, the word it uses is velocity. Confusing, unnecessarily.

    One thing I have learned over the years is that almost all exercises have a common characteristic: they are doable. And most of the time they ask for skills and knowledge recently treated in book or class. Sometimes that helps me to turn the question around: given what I am supposed to have learned, what can the author of the exercise possibly want from me ?

    In this case I find it hard to believe the expression for "distance covered under uniform accelerated motion", that is clearly needed for this exercise, hasn't been treated. If that is really so, the exercise is premature and you are excused.

    Again: best of 'luck' tomorrow !
    BvU

    PS no snarkiness (I didn't even know the word!) intended. That's the nice thing in forums: what you don't like you can ignore and shed off like a duck sheds water.
     
    Last edited: Feb 5, 2015
  23. Feb 5, 2015 #22

    NascentOxygen

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    I think you should get yourself a suitable textbook, and read through it on your own initiative. You need a solid reference. A lot of textbooks are written with self-study in mind, because teacher qualifications can be patchy.
     
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