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Equations of Motion (Deriving equations)

  1. Oct 30, 2007 #1
    Hi, I am just wondering how you would approach this problem:
    Using the definitions below, derive an equation for velocity as a function of acceleration and time (v=f(a,t)). Assume initial velocity is Vo. The answer to this problem is v=v0+a(t2-t1). My question is how would you arrive to this answer step-by-step. Below are the definitions:

    x=current position in the x dimension
    deltax= change in position
    t=time now, t0 is the starting time.
    deltat= a time interval, t2-t1.
    v=deltax/deltat (use as a scaler for now).
    deltav= a change in velocity.
    a=deltav/deltat (Use as a scaler for now).

    Subscripts: 0 is an initial value, other numbers are subsequent values in time order as needed.

    v (average)= (v1+v2)/2, a simple average.

    Any help will be appreciated. Thank you.
  2. jcsd
  3. Oct 30, 2007 #2
    # I would first look at the definitaion of a
    [tex] a(t) = \frac{dv(t)}{dt} [/tex]
    # Integrate both sides
    [tex] \int^{t}_{t_0} a(t) dt = v(t) - v(0) [/tex]
    # This is the relation between v and a. but it is not in the form of you want, i.e. [tex]v=f(a,t)[/tex]. It is an integral equation.
    # So, there should be an assumption about a, which changes integral the relation between a and v to a function. I think if you think, you can find it by yourself.
    Last edited: Oct 30, 2007
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