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!