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cheer_chic
Sep3-03, 02:39 PM
We are supposed to derive an equation that solves for displacement that does not use time. We can use these two equations. d(t) = vit + 1/2 at squared. or d(t) = 1/2 (vit + vft) t
It's due tomorrow. Any help would be greatly appreciated!
Tom Mattson
Sep3-03, 04:25 PM
Should be a piece o'cake. You have:
d=vit+(1/2)at2
and you also have:
vf=vi+at.
Try to combine those to get the equation you are looking for. If you get stuck, show us how you started and we will help you from there.
Inquiring_Mike
Sep3-03, 04:34 PM
I'm not sure if I decoded the equation lines very well... but i think this is how you do it.
Take the second equation and solve for time and then substitute that equation into the first equation and that will get rid of the time.
If i'm not helping at all, watch.... (^2 is squared)
d= 1/2 (vit + vft) t (expand t)
= 1/2 (vit^2 + vft^2) (pull out t^2)
= 1/2 t^2 (vi + vf) (multiply both sides by 2)
2d = t^2 (vi +vf) (some rearranging)
t= the square root of (2d /vi+vf)
Now, wherever there is a t in the first equation, substitute the equation that you found...
:)
I hope that is right ......
Inquiring_Mike
Sep3-03, 04:36 PM
Tom, I'm not sure that she is allowed to use any other formulas either than the two she is given.... Do you agree with my work?
I assume vi = v(intial) and vf = v(final), usually v(initial) is denoted as 'u' and v(final) as 'v'.
It's a very simple equation, it's a simple case of re-arranging one of the equations and substitung it into the other, you should be able to do it yourself but for your reference the answer is:
d = (v2/2 - uv)/a
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