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Point slope equation and velocity

  1. Feb 19, 2012 #1
    1. The problem statement, all variables and given/known data

    Is it correct that the point slope equation describing a line corresponds to the position formula in physics?

    can i say: y-y=m(x-x) is equivelent to y-y=v(t-t)

    so that y=mx+y is equivelent to y=vt+y or y(t)=y + vt

    and if I take the derivative of y(t), I get y'(t) = v or just v(t) = v
  2. jcsd
  3. Feb 19, 2012 #2
    That's true if there is no acceleration involved. You have described the equation of a line [tex]y=mx+c[/tex] At constant velocity (ie. no acceleration) there will be a linear relationship between position and time and the slope of this line (the derivative of position with respect to time) will be the velocity.

    The full form of this kinematic equation is [tex]s=s_0+vt+\frac{1}{2}at^2[/tex] where [itex]s_0[/itex] is the initial position.

    You can see that when [itex]a=0[/itex] this reduces to [tex]s=vt+s_0[/tex]which is the equation of a line.

    When you differentiate this w.r.t. time you get that[tex]s'(t)=v[/tex]Hope this helps :)
  4. Feb 19, 2012 #3
    Hey GothFraex, yes, it does help, very much.

    For a kinematic equation with acceleration [tex]s=s_0+vt+\frac{1}{2}at^2[/tex] is essentially the quadratic equation f(x) = ax^2+bx+c and since taking the derivative will be a line of the general form y = mx + b, you'd get [tex]v(t)=v_0+at[/tex] and that has the same graph form as s = vt + s[itex]_{0}[/itex]

    So the graph of position vs time without acceleration is the same equation as velocity vs time with acceleration? And the difference in their meaning is the axis.

    Is this correct also?
  5. Feb 19, 2012 #4
    Yes this is correct.

    There is a linear relationship between position and time for constant velocity (the slope of the graph is constant).

    Similarly there is a linear relationship between velocity and time for constant acceleration (again since the slope is constant).

    By the way, the familiar kinematic equations are always for constant acceleration.
  6. Feb 19, 2012 #5
    Okay, because it starts to get a little confusing when it comes to breaking down projectile motion.

    For instance, projectile motion in two dimensions are usually illustrated as a parabola on a position vs position graph. However, the actual equations used to solve questions are not even based on that graph. x(t) = x +vt + .5at^2 is a graph that would be based on a position vs time graph, right? So the horizontal x component of position could be expressed on its own seperate p vs t graph, and the vertical y component of position(t) could be expressed on its own seperate p vs t graph too, right?
  7. Feb 20, 2012 #6
    Yes that's true.

    When you work with a 2D projectile motion problem you break it up into components so in a sense it is like working with 2 different position versus time graphs.

    The kinematic equations still describe the motion, however, you now have to deal with vectors.
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