- #26

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quadratic formula: t= 1.13814 E -8 or t=1.19729 E -8

Using vf = vi - a*t: t= 1.16771 E -8

I keep redoing the math but it turns out the same. :/

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- Thread starter C6ZR1
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- #26

- 54

- 0

quadratic formula: t= 1.13814 E -8 or t=1.19729 E -8

Using vf = vi - a*t: t= 1.16771 E -8

I keep redoing the math but it turns out the same. :/

- #27

gneill

Mentor

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quadratic formula: t= 1.13814 E -8 or t=1.19729 E -8

Using vf = vi - a*t: t= 1.16771 E -8

I keep redoing the math but it turns out the same. :/

This is an example of where significant figures and rounding can have marked effects on the results. ALWAYS calculate intermediate values using more decimal places than the final result requires, and round only at the end. Quadratics can be notoriously sensitive to parameter values. This particular one

However. Let me give you a tip that may help. When you are working on a problem that has some initial velocity v

[itex] d = v_o t - \frac{1}{2} a t^2 [/itex]

Use instead:

[itex] d = \frac{1}{2} a t^2 [/itex]

which provides: [itex] t = \sqrt{\frac{2 d}{a}} [/itex]

and thus only one root to contend with.

The thing is, going from some initial velocity to zero over distance d is just a time-reversed version of going from zero to that speed over the same distance! It's like "running the film backwards" -- the event takes the same amount of time to complete only the formula for the distance is simpler, dropping the initial velocity term.

- #28

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