# Constant acceleration problem to find the speed of a rocket

1. Jan 14, 2013

### mahrap

A rocket starts from rest and moves upward from the surface of the earth. For the first 10 of its motion, the vertical acceleration of the rocket is given by 2.60t, where the -direction is upward.

What is the speed of the rocket when it is 240 above the surface of the earth?

v = v0 + at
change in x = (v + v0)t/2

I solved for t in the second equation by plugging in 240 for change in x, 0 for v0, and left v as v: meaning i solved for t in terms of v. Then i took this and plugged it in the t of the first equation with t=480/v, a=2.6(480/v) and v0=0. However i keep getting a wrong answer. What should i do?

2. Jan 14, 2013

### Staff: Mentor

The acceleration is not constant, so the usual kinematic equations for constant acceleration will not apply. I see an integral or two in your future

3. Jan 14, 2013

### mahrap

What would be the limits of my integration. How would I find the time where the rocket is at the given height to integrate the acceleration to get velocity?

4. Jan 14, 2013

### Staff: Mentor

Integrate to find expressions for velocity as a function of time and distance as a function of time. Then use these expressions to solve the particular problem.

As for the limits, well what variable are you integrating over?