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...

[tex]\int_{0}^{19} {19c} dx + \int_{19}^{23} {-39.2} dx = 4700[/tex]

grrrr. nope. Also sig figs matter.
I got a 13.5 answer...and a 13.2 answer...this is getting fustrating.

 

[kyle8921]

Your approach is on the right track, but there are a couple of errors in your calculations and approach. Let's break it down step by step:

(a) To find the rocket's acceleration during the first 19 seconds, we can use the formula a = (vf - vi)/t, where vf is the final velocity, vi is the initial velocity, and t is the time interval. In this case, we know that the rocket starts from rest (vi = 0) and reaches a final altitude of 4700 m in 23 seconds (t = 23). So, the acceleration during the first 19 seconds would be:

a = (4700 m - 0 m)/(23 s - 0 s) = 204.3 m/s^2

(b) To find the rocket's speed as it passes through a cloud 4700 m above the ground, we can use the formula v^2 = vi^2 + 2ad, where vi is the initial velocity, a is the acceleration, and d is the displacement. In this case, we know that the rocket's initial velocity is 0 m/s, the acceleration is the same as before (204.3 m/s^2), and the displacement is 4700 m. So, the rocket's speed as it passes through the cloud would be:

v = √(0 m/s)^2 + 2(204.3 m/s^2)(4700 m) = 1021.8 m/s

Your final distance formula is not quite correct, as it mixes up the two different time intervals (0-19 seconds and 19-23 seconds). To find the rocket's final altitude, we need to use the formula d = vit + 0.5at^2, where vi is the initial velocity, a is the acceleration, and t is the time interval. In this case, we can break it down into two parts:

- For the first 19 seconds, we can use the formula d = 0.5at^2, where a is the acceleration we calculated in part (a) (204.3 m/s^2) and t is the time interval (19 seconds). This gives us a displacement of:

d = 0.5(204.3 m/s^2)(19 s)^2 = 180.5 m

- For the remaining 23-19 = 4 seconds, we can use