Can someone explain the confusion with these basic motion equations?

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FlamingAero
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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.
 
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FlamingAero said:
Are these values not compatible with this formula?
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.)
 
Doc Al said:
Your values are just physically impossible.

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