# Particle motion question

1. Sep 24, 2006

### Robokapp

okay...i labeled acceleration 0 -19 as c.

a=c
v=cx
d=0.5cx^2

for 19-23 we have

a=-9.8
v=-9.8x
c=-4.9x^2

0 -19 we have d=180.5c
19-23
we have -823.5

180.5c-823.5=4700

Am I doing it right in solving for C which is constant acceleration in 0 -19?

edit: ok I know now it's not right.

my final distance formula would be

4700=-4.9x^2(for 19</x</23)+0.5cx^2(0</x</19)

But i dont know how to put this in a real equation form...

Last edited: Sep 24, 2006
2. Sep 24, 2006

### Robokapp

my second approach...

v=cx-9.8x

v=19c-9.8(4)=19c-39.2

and now i take the integral of all this and set it equal to 4700

okay...so i think the following formula should work...

$$\int_{0}^{19} {19c} dx + \int_{19}^{23} {-39.2} dx = 4700$$

grrrr. nope. Also sig figs matter.