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Basic Motion Equations

  1. May 30, 2013 #1
    I have been reviewing the basic two-dimensional motion equations and I've discovered a conundrum that is causing me much confusion. For example, here is a basic formula with variables:

    [itex]v^2 = vi^2 + 2ax[/itex]

    [itex]v = ?[/itex]

    [itex]vi = 27[/itex]

    [itex]a = -7.5[/itex]

    [itex]x = 49[/itex]

    Therefore:

    [itex]v^2 = 27^2 + 2(-7.5)(49)[/itex]

    [itex]v^2 = 729 + -735[/itex]

    [itex]v = √(-6)[/itex]

    When I input the square root of (-6) into my calculator (a TI-83+), I receive a ERR:NONREAL ANS message. Are these values not compatible with this formula?

    Here's another similar example, this time with the formula:

    [itex]ΔX = vi*t + (1/2)at^2[/itex]

    [itex]ΔX = 49[/itex]

    [itex]vi = 27[/itex]

    [itex]a = -7.5[/itex]

    [itex]t = ?[/itex]

    I have no idea how to even arrange the equation in terms of [itex]t[/itex]. Is this formula limited to solving displacement?

    Thank you for your help and guidance.
     
  2. jcsd
  3. May 30, 2013 #2

    Doc Al

    User Avatar

    Staff: Mentor

    Your values are just physically impossible. Given that initial velocity and acceleration, you'll never achieve x = 49. (Figure out the maximum value of x.)

    Similar issue with the other formula for time. (In general, you can surely solve for the time. You'll get a quadratic equation.)
     
  4. May 30, 2013 #3
    I now see my error. The value [itex]ΔX = 49[/itex] was rounded for significant figures, and should have instead been [itex]ΔX = 48.6[/itex]

    Thank you for your help.
     
    Last edited: May 30, 2013
  5. May 30, 2013 #4
    Yeah the second one is a quadratic so you can either set it to 0 and factorise to get your two answers or use the quadratic formula below

    [tex]

    \frac{-b\pm\sqrt{b^2-4ac}}{2a}
    [/tex]
     
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