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Kinematics, deriving equations.

  1. Jan 30, 2017 #1
    1. The problem statement, all variables and given/known data
    "Derive the equations for position (in terms of acceleration, initial position, initial velocity, and time) and velocity (in terms of constant acceleration, a, initial velocity, v0, and time, t) from the definitions of position, velocity, and acceleration (derivative definitions).

    2. Relevant equations
    x = v0t + (at2)/2
    vf = vi + at
    x = [(vf + vi) / 2]t
    vf2 = vi2 + 2ax
    d/dx[x(t)] = v(t)
    d/dx[v(t)] = a(t)
    xf = xi + ∫ 0tvdt

    3. The attempt at solution
    I don't know exactly what it is asking. I got confused by the very end... Thanks!
     
  2. jcsd
  3. Jan 31, 2017 #2
    Hmmm. I'm not sure how to give a hint without pretty much giving it away. Start with your second to last equation and assume acceleration is constant. I think you can tell what to do. (By the way, that's dt, not dx.)

    I think some confusion may arise from familiarity. This derivation is how they describe these things in class and right at the front of the book. It seems so basic and fundamental that you have a hard time thinking of it as an answer to a question. Surely it's a starting point not the goal. Surely they want more than this. Nope. That's all they want. They want you to derive your equation 1 and equation 2 starting with your equations 5 and 6 (after you correct them) just like they do in chapter 1.
     
  4. Jan 31, 2017 #3
    You have the answer in your relevant equations.

    We say that change in position is the velocity.

    That is:
    \begin{equation}
    v(t) = \frac{\Delta x}{\Delta t}
    \end{equation}
    We used this notation in alegebra-based physics because we are taking the difference between two points which is a secant line (i.e rise/run).

    Luckily, you know calculus, where you can find the instantaneous rate of change which is given by a tangent line.

    Therefore, our velocity equation becomes:
    \begin{equation}
    v(t) = \frac{dx}{dt}
    \end{equation}

    You know that the change in velocity is the acceleration, that is:
    \begin{equation}
    a(t) = \frac{dv}{dt}
    \end{equation}
    If a = constant.

    Integrate a(t) to get v(t)
    If a is a constant, then the integral is simple.
    \begin{equation}
    v(t) = \int_{t_1}^{t_2} a \ dt = a \int_{t_1}^{t_2} dt = at + c
    \end{equation}

    but the constant c is in this case, v$_0$ (you're starting velocity)

    Follow the logic for the rest of the problem. Keep integrating until you get to position (i.e x(t).)
     
  5. Jan 31, 2017 #4
    Wow, okay, I am sort of seeing what the problem is asking. Thanks! I thought for a moment that I needed to derive specific kinematic equations...
     
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