Drageur Veteran 125 Posts user info edit post |
...if not all of you. but i am a dumb ass (also can you explain how you did it?)
A model rocket accelerates upward from the ground with a constant acceleration, reaching a height of 91 m in 4.2 s.
What is the acceleration (in m/s 2)? What is the speed (in m/s) at a height of 91 m? 8/31/2006 5:06:25 PM |
tjoshea All American 4906 Posts user info edit post |
x = (1/2) a t^2 v = a t 8/31/2006 5:09:40 PM |
amac884 All American 25609 Posts user info edit post |
101101110110001 8/31/2006 5:10:57 PM |
Drageur Veteran 125 Posts user info edit post |
^^ i cant use that first one cause i dont know a...right? 8/31/2006 5:13:57 PM |
Førte All American 23525 Posts user info edit post |
e^(pi * i) 8/31/2006 5:19:14 PM |
loudRyan All American 594 Posts user info edit post |
^^ you know x and t, so you can solve for a 8/31/2006 5:29:41 PM |
BigDave41 All American 1301 Posts user info edit post |
^^^thats the point. you don't know "a", so you find an equation with variables that you know and use that to solve for the unknown.
i really enjoyed all the kinematics stuff from ap physics. this brought me back.
[Edited on August 31, 2006 at 6:06 PM. Reason : .] 8/31/2006 6:06:06 PM |
Drageur Veteran 125 Posts user info edit post |
Thanks for the help all. 8/31/2006 6:07:57 PM |
darkone (\/) (;,,,;) (\/) 11610 Posts user info edit post |
I'll help you...
If A is constant, integrate to get velocity:
A(t) = A %A is a constant (acceleration)
V(t) = At + V0 %V0 is initial velocity (zero in this case)
Integrate again to get position:
D(t)= (1/2)A(t^2) +V0t + D0 %Do in initial displacement (zero in this case)
t in all cases is time
Since the rocket started from rest at the ground V at time zero, or V(0)=0
In fact, you don't really need all that math to solve this problem.
Since acceleration is constant and you started from rest, your average speed is one half your final speed. (work with the above equations to see why) Your average speed is your [total displacement / net time]. Double that value for your speed at time 4.2s. Once you know your final speed you can calculate your acceleration. Acceleration is your cange in speed per unit time. Thus, your acceleration in this case is your [final velocity / time it took to get to the final point].
If you can't solve the problem from what I've given you, drop out of college. (not a joke, just trying to save your parents money) 8/31/2006 6:19:26 PM |
1 All American 2599 Posts user info edit post |
this isn't rocket science 9/1/2006 9:50:47 AM |
darkone (\/) (;,,,;) (\/) 11610 Posts user info edit post |
^ You come in a day late to make a bad joke.... go back to chit-chat. 9/1/2006 10:45:12 AM |
1 All American 2599 Posts user info edit post |
tjoshea answered the question in the second post to the thread 9/1/2006 12:00:49 PM |
darkone (\/) (;,,,;) (\/) 11610 Posts user info edit post |
^ and the kid obviously understood the answer 9/1/2006 3:52:19 PM |
Jere Suspended 4838 Posts user info edit post |
Quote : | "^^ i cant use that first one cause i dont know a...right?" |
jesus christ9/2/2006 11:47:33 AM |