What is the rocket's final altitude 23 seconds after launch?

[Robokapp]

A 1000 kg weather rocket is launched straight up. The rocket motor provides a constant acceleration for 19 s, then the motor stops. The rocket altitude 23 s after launch is 4700 m. You can ignore any effects of air resistance.

(a) What was the rocket's acceleration during the first 19 s?


b) What is the rocket's speed as it passes through a cloud 4700 m above the ground?

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 don't know how to put this in a real equation form...

 

[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.
I got a 13.5 answer...and a 13.2 answer...this is getting fustrating.