PLEASE help on this simple problem.

[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).

:cry:

[Ba]

The 1000 kgs seems to suggest some force analysis.

[FancyNut]

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

[ehild]

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?



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).

:cry:

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

[FancyNut]

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


*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... :redface:

[ehild]

*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... :redface:


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

ehild