Tuesday

October 21, 2014

October 21, 2014

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

**Answer this Question**

**Related Questions**

physics - Motion in a Circle A particle P travels with constant speed in a ...

Physics - A cat rides a merry-go-round turning with uniform circular motion. At ...

Physics - A cat rides a merry-go-round turning with uniform circular motion. At ...

physics - A 10 g particle undergoes Simple Harmonic Motion with an amplitude of ...

physics help - The magnitude of the velocity of a particle which starts from ...

physics help - The magnitude of the velocity of a particle which starts from ...

Physics - A particle P travels with constant speed in a circle of radius 7.3 m ...

Physics - A particle P travels with constant speed in a circle of radius 7.3 m ...

Physics - What quantitative information do you need to calculate each of the ...

Physics/Math - At time t1 = 2.00 s, the acceleration of a particle in ...