# Kinematics and finding acceleration

1. Oct 2, 2013

### JohnCy

1. The problem statement, all variables and given/known data

A weather rocket is launched straight upwards. The motor provides a constant acceleration for
18s, then the motor stops. The rocket altitude after 26s after launch is 5800m. What was the rocket's acceleration during the first 18 seconds?

2. Relevant equations
[ tex ]x = x_0 + v_0 t + (1/2) a t^2
[/tex]
[ tex]v = v_0 + a t
[/tex]

3. The attempt at a solution
This answer doesn't seem right to me. I wasn't confident working it out.

5800=.5(a)(18)^2-(.5)(9.81)(8)^2
162a=5800+313.92
162a=6113.92
a=37.7

2. Oct 2, 2013

### PhanthomJay

You forgot to include the initial velocity of the rocket during the second part of the trip with the motor stopped.

3. Oct 3, 2013

### JohnCy

I don't know the velocity at 18 seconds. All the initial values are given, they are all zero. The only intermediate value given is 18 seconds. That's why I'm having a difficult time, I'm not sure how to handle all the unknowns. I'm trying to draw some equations from the three graphs I drew but some of the equations don't make sense. Thank you for your reply.

4. Oct 3, 2013

### PhanthomJay

you can use your 2nd relevant equation , where v in that equation after 18 s becomes v_o in your first relevant equation . You will have to leave a as unknown, then solve.

5. Oct 3, 2013

### HallsofIvy

Staff Emeritus
Let the "constant acceleration" while the rocket engine is firing be "a". Then the height when the engine stops firing, from s= (a/2)t^2, is (a/2)(18)^2= 162a meters and its speed is 18a meters per second. After that, we have only the acceleration due to gravity: h= 162a+ 18a(t- 18)- 4.9(t- 18)^2. (Do you see why "t- 18"?) Set t= 26 s, h= 5800 m and solve for a.

6. Oct 3, 2013

### JohnCy

Thanks for both your help guys. I've been doing physics for all of three weeks now and it doesn't seem to be getting easier. Sorry for the messy equations. I tried to use latex but apparently it didn't work. Thanks again!