# Homework Help: A rocket going up with acceleration =

1. Nov 10, 2015

### David112234

1. The problem statement, all variables and given/known data
A rocket starts from rest and moves upward from the surface of the earth. For the first 10.0 s of its motion, the vertical acceleration of the rocket is given by ay=(2.90m/s^3)t, where the +y-direction is upward.

What is the height of the rocket above the surface of the earth at t = 10.0 s ?
Express your answer with the appropriate units.

2. Relevant equations
v=dy/dt
a=dv/dt
y =v*t + (1/2) a*t^2
3. The attempt at a solution

I drew a diagram. A rocker going up, from 0 -10 a= 2.90 *t
from 10 and higher, a = -9.802
I integrated the acceleration from 0-10, and got the velocity to be 2.90/2 t^2 I used the kinematics formula for position, height in this case and got
h= 2.90/2 t^3 + .5*-9.802*t^2 = 959.9
its not correct, what am I doing wrong? What should I do instead and why?

2. Nov 10, 2015

### SteamKing

Staff Emeritus
Not sure why you care what happens to the rocket for t > 10 sec. The problem asks you to find the height of the rocket at t = 10 sec.

What happens to the rocket after that is immaterial.

You only need to concern yourself with what happens to the rocket over the interval 0 ≤ t ≤ 10.

If you can integrate acceleration once to find velocity, why can't you integrate velocity to find height?

3. Nov 10, 2015

### David112234

I'm sorry, I didn't not read carefully enough, Im pretty sleepy doing tons of home work, I did a previous question that asked for something past the original time interval, and I was in the same mind set with this one. Thanks!