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January 27, 2015

January 27, 2015

Posted by **karan** on Tuesday, August 19, 2014 at 3:38pm.

- physics -
**Damon**, Tuesday, August 19, 2014 at 5:26pmit evidently goes forward a distance 3d then back a distance 2d to end up a displacoutup to be negative.

deceleration phase out to 3 d

t1 = time out to 3 d

v = Vi + a t1

0 = Vi + a t1

so

t1 = -Vi/a

3 d = Vi t1 + (1/2) a t1^2

3d = -Vi^2/a + .5 Vi^2/a = -.5Vi^2/a

d a = -(1/6) Vi^2

return phase back from 3 d to one d

Vf = 0 + a t2

so

t2 = Vf/a

-2d = (1/2) a t2^2

-2d = .5 Vf^2/a

--------------------------------

algebra from here on :)

d a = -(1/4) Vf^2

(1/6) Vi^2 = (1/4)Vf^2

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