(adsbygoogle = window.adsbygoogle || []).push({}); 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 dont know how to put this in a real equation form...

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