1. Oct 21, 2004

FancyNut

A 1000kg weather rocket is launched straight up. The rocket motor provides a constant acceleration for 16 s, then the motor stops. The rocket altitude 20 s after launch is 5100 m. You can ignore any effects of air resistance.

What's the rocket's acceleration during the first 16 seconds?

I posted this in another thread but I guess nobody is looking at it... if posting this is against the rules then I guess ban me/lock thread but PLEASE how do I even start? Initial velocity is zero, I know time, but with no displacement (the 5100 is for the very end after the motor stopped) or final velocity (when the motor stops) I just can't get it. I used all 3 equations and the best I can do is get acceleration in terms of final postion (the one where the motor stops).

2. Oct 21, 2004

Ba

The 1000 kgs seems to suggest some force analysis.

3. Oct 21, 2004

FancyNut

It's from a chapter before forces are introduced so it shouldn't have anything with the problem...

4. Oct 21, 2004

ehild

Try to think logically instead of plugging in data into formulas.... What happened to the rocket? It rose with constant acceleration for t1=16 s, after that the motor stopped but gravity still acted on it, so it rose with deceleration of g for t2=4 s.
You do not know the acceleration during the first t1 time, so it is "a".
The velocity rose to v1=a*t1, and the displacement is h1=a/2*t1^2.
Now the second period comes, for t2=4 seconds, with acceleration -g and "starting" velocity vo=v1. The displacement is h2=vo*t2 -g/2 *t2^2= a*t1*t2-g/2*t2^2.
The sum of both displacements is h = h1+h2 =5100 m. You have one equation with one unknown, it is the acceleration, a.

ehild

5. Oct 21, 2004

FancyNut

*gives ehild a box of cookies*

:D :D :D

I guess next time I shouldn't be so lazy and just expect to plug in numbers in a given formula...

6. Oct 22, 2004

ehild

Great!!!! and thanks for the virtual cookies. They were yummy :rofl:

ehild