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Basic dynamics problem

  1. Sep 18, 2014 #1
    1. The problem statement, all variables and given/known data

    I've attached a picture of the problem

    2. Relevant equations

    I'm assuming I need to use ads=vdv

    3. The attempt at a solution

    I've attached a picture of my attempt at solving the problem, but I really have no idea what I'm doing. I'm really having a hard time in this class because my professor only ever goes over theory and proofs. It doesn't help that it's been 5 years since I took calculus.
    1. The problem statement, all variables and given/known data



    2. Relevant equations



    3. The attempt at a solution
     

    Attached Files:

  2. jcsd
  3. Sep 18, 2014 #2

    SteamKing

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    The image showing the problem statement is barely legible. Can you take another picture and repost please?
     
  4. Sep 18, 2014 #3

    BiGyElLoWhAt

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    Where did you get that equation?

    If you know calculus, why don't you try to derive a vector position function P(t)-><x,y> using your acceleration and initial conditions. You can take the ground to be y = 0 (or wherever really, but watch your heights) and the left side of the cliff to be x = 0. You have a constant acceleration, g, so it should be relatively easy to figure out.
     
  5. Sep 18, 2014 #4

    BiGyElLoWhAt

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    @steamking.

    Dude standing at edge. First cliff is 200' high, second cliff is 180' high. If he jumps with an initial velocity 8ft/s ihat how far can the cliff be if he wants to not fall to his death.
     
    Last edited: Sep 18, 2014
  6. Sep 18, 2014 #5

    BiGyElLoWhAt

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    Also to finish my thought (oops haha) :

    With that position function, you can solve for the x value where y = a certain height (i.e. the height of the second cliff)
     
  7. Sep 18, 2014 #6
    Here's a better picture of the problem, sorry
     

    Attached Files:

  8. Sep 18, 2014 #7
    I haven't taken calculus in 5 years, so I don't really think I would say I know calculus haha how would I derive the vector position function you're talking about?
     
  9. Sep 18, 2014 #8

    gneill

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    The usual kinematic formulas should suffice for this. No calculus required.

    Hint: His initial velocity is horizontal. What does that tell you about his initial vertical velocity? How long does it take him to fall 20 ft (200 - 180)?
     
  10. Sep 18, 2014 #9
    I think I've solved it with the kinematics equations, but the problem is part of a practice test and my professor wants me to show my work using calculus for all problems. I'm so confused.
     
  11. Sep 18, 2014 #10
    The gentleman (judging by the hat) must move by a distance d at the exact same moment that he reaches high h2.

    So, you have to find the time t required for him to move by a distance h1-h2 horizontally.

    Then just plug this time on the equation for horizontal motion of the gentleman. (the motion is decomposed in two different motions along y and x axis)
     
    Last edited: Sep 18, 2014
  12. Sep 18, 2014 #11
    does this look right?
     

    Attached Files:

  13. Sep 18, 2014 #12
    I think the only calculus you could apply is in the derivation of the motion equations (as
    BiGyElLoWhAt said). For that you need the "calculus definition" of the concepts of velocity and accelaration, which, since you professor focus too much on theory, I believe you've covered on class (if not check the introductory part of any calculus-based physics textbook).
     
  14. Sep 18, 2014 #13
    the motion is decomposed into two different motions: on along X and the other along Y. pay attention to what type of motion each one is.
     
  15. Sep 18, 2014 #14

    BiGyElLoWhAt

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    Not so much.

    Whats the velocity vector at t =0 ? (the instant he jumps)

    Whats the acceleration vector at t = 0?

    What's the position vector at t= 0?

    How do these three quantities relate to each other? What can I do with P(t) to get to V(t)? what can I do with V(t) to get to A(t)? How can I reverse these processes?
     
  16. Sep 18, 2014 #15
    I don't know what you mean
     
  17. Sep 18, 2014 #16
    I just don't get what to do without time or the final x position
     
  18. Sep 19, 2014 #17

    BiGyElLoWhAt

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    Ok, basic calculus lesson.
    Velocity is the rate of change of position with respect to time meaning ##v(t) = \frac{d}{dt}P(t)##
    Acceleration is the rate of change of velocity with respect to time meaning ##a(t)=\frac{d}{dt}v(t) = \frac{d^2}{dt^2}P(t)##

    You can reverse these processes by calculating the antiderivative (basically the derivative operation but in reverse, it differs a little for more complex functions)

    If you have an acceleration, in this case g, you can write that as a vector. Which way does gravity pull on objects in terms of unit vectors? (or whatevery notation you prefer) <x,y,z> ##x\hat{i} + y\hat{j} + z\hat{k}##

    When you calculate the antiderivative, you end up with a constant of integration (when you take the derivative of a constant, you get 0, so you have to add a constant into your function) and you need to use boundery conditions to solve for that constant. So when you do this problem, you start with the acceleration vector, and you move to the velocity vector by way of antiderivatives. You take some kind of condition that you know the value of the velocity for and use that to solve for your constant. So in this case you know the initial velocity, that's how fast sir hat wearer jumps. The cool thing about this problem is that you know the initial velocity at t=0, or the beginning of your motion you're trying to model. So ##v(t) = \int a(t)dt \ \ \text{&}\ \ v(0) = v_0 \text{(initial velocity)}## See if you can calculate this. Post back what you have.
    If you need a refresher, I personally like Paul's notes: http://tutorial.math.lamar.edu/Classes/CalcI/IndefiniteIntegrals.aspx

    Also, this might be a better page, but ultimately I don't really know what you know/don't know, so you might be the best person to gather resources.
    http://tutorial.math.lamar.edu/Classes/CalcIII/Velocity_Acceleration.aspx
     
    Last edited: Sep 19, 2014
  19. Sep 20, 2014 #18
    Do you need calculus ?
    The core of the problem is the the time it takes to fall vertically 20 feet (solvable with newtons rules and a value for g), then use the time to solve the width.
     
  20. Sep 20, 2014 #19
    20140920_181025.jpg
    thanks for the help @BiGyElLoWhAt. I dont know why I was having such a difficult time with this problem.
     
  21. Sep 20, 2014 #20
    20140920_181706.jpg
    @BiGyElLoWhAt I was hoping you might take a look at this problem and see if it looks right. According to the answer key my professor sent out today, the velocity is correct but the acceleration is not. My professor has a=80.04m/s^2
     
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