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A (hopefully) original theorem

  1. Apr 30, 2015 #1
    Hello fellow members of PF,

    I believe I may have created a theorem for calculating the distance travlled on the 2-dimensional plane when gravity and other forces are ignored. Now, if this has been thought of before, that's not my fault. I have not studied this before.

    The theorem is this... the distance travelled along the Y axis is equal to the gradient of travel divided by 90, then multiply the speed(m/s) by the time taken (s) (yes, this is d=vt), and then multiply both those terms together.

    Then, to also find the distance travelled along the X axis, take v times t and then subtract from it the distance travelled along Y. Sounds fiddley in sentences. But I have enclosed a whiteboard sketch.

    So, is this original? Does it work? Has it been thought of before?


    g/90 x vt = distance moved in y, where g = gradient , angle.

    motion along x = (vt) - (g/90 x vt)
  2. jcsd
  3. Apr 30, 2015 #2


    Staff: Mentor

    Your problem is that of projectile motion and you can determine if your nethod works by comparing it to how its done in introductory physics:


    Does your solution when plotted on the xy plane produce a parabolic arc?
  4. Apr 30, 2015 #3


    User Avatar
    Homework Helper

    The OP is ignoring gravity.

    No, it's wrong.

    What you've done is to complicate simple trigonometry.

    Given a right triangle that sits on the x-axis and the hypotenuse is the distance (d) and [itex]\theta[/itex] is the angle that you're travelling at (where [itex]\theta=0[/itex] is along the x-axis), then the distances x and y are


    You can also replace d by vt if you wish, since d=vt. Also, [itex]\theta[/itex] is in radians. If you use degrees (g, to use your parameters) instead then you'll have to make the transformation

    [tex]\cos\theta = \cos{\frac{\pi g}{180}}[/tex]

    So finally you'll have

    [tex]x=vt\cos{\frac{\pi g}{180}}[/tex]
    [tex]y=vt\sin{\frac{\pi g}{180}}[/tex]

    Now, g/90 may be incorrect, but it's not too bad of an approximation (for values between 0-90 degrees, but if you used more than that then it's way off), but a better approximation (which is just approximating sin(x) and cos(x) at [itex]x=\pi/4[/itex] radians, or 45 degrees) would be

    [tex]\sin\frac{\pi g}{180}\approx \frac{1}{\sqrt{2}}\left(1+\pi\left(\frac{g-45}{180}\right)-\pi^2\left(\frac{g-45}{180}\right)^2\right)[/tex]
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