# Kinematics and finding acceleration

JohnCy

## Homework Statement

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?

## Homework Equations

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

## 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

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

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.