# Homework Help: 1D motion

1. Sep 23, 2011

### barthayn

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

A 1000 kg weather rocket is launched straight up. The rocket motor provides constant acceleration for 16 s, then the motor stops. The rocket altitude 20 s after launch is 5100 m. You can ignore any effects of air resistance.

a) What was the rocket's acceleration during the first 16 s?
b) What is the rocket's speed as it passes through a cloud 5100 m above the ground?

t = 16s

2. Relevant equations

None that I know :(

3. The attempt at a solution

I assumed that at its highest point it stop accelerating and fell back down to the earth. I got 223.56 m/s from the height of 5100 m and the acceleration to gravity. I believe this is wrong, but it was the only thing to go on. Do I used this formula, 0 = 223.56t + 0.5at, where t = 16. Solved for a and got - 447.12m/s/s. This is clearly wrong because I made up positive and down negative. This is claiming it is falling at a rate of 447.12m/s/s downwards. Any hints on where to begin?

Last edited: Sep 23, 2011
2. Sep 23, 2011

### Ignea_unda

Firstly, can you provide the question? The problem statement you gave seems to be missing what we are seeking.

3. Sep 23, 2011

### barthayn

Sorry about that, I was so sidetracked about how to figure out where to begin I forgot to include the question (Lol). Anyways, if you haven't notice any change in my first post it is as follows:

A 1000 kg weather rocket is launched straight up. The rocket motor provides constant acceleration for 16 s, then the motor stops. The rocket altitude 20 s after launch is 5100 m. You can ignore any effects of air resistance.

a) What was the rocket's acceleration during the first 16 s?
b) What is the rocket's speed as it passes through a cloud 5100 m above the ground?

4. Sep 23, 2011

### Ignea_unda

Look at the problem in stages, with a Free Body Diagram for each of the stages.

For the first stage, you have information from 0s to 16s. On the second stage you have information from 16s to 20s. There's only one difference between the two stages. Put in your forces and your mass. Then you can solve for your acceleration. Note that the rocket isn't necessarily stopped at 5100 m. It's accelerating the entire time there is thrust, so there will still be velocity at the end. Let me know if this helps you at all.

5. Sep 23, 2011

### barthayn

To calculate the constant acceleration during the first 16 s I think about forces, however, we do not have the applied force. So I concluded that there is no way to calculate the net force on the object, and therefore, cannot calculate the acceleration of the object. Or am I missing a key piece of information here?

6. Sep 23, 2011

### Ignea_unda

Re-reading the problem, it looks like I fell into the pitfall of extra information, my apologies. We don't need forces at all. Sorry for misleading you.

The key piece of information is that you need to look at both stages together to solve the problem. Your acceleration for the first part is "a" plus gravity (since gravity is down, it'd be a negative acceleration). The second part you lose that acceleration "a" and only have gravity. At the end of 20s, you know where the rocket is. Think about how you get position from acceleration.

7. Sep 23, 2011

### barthayn

Are you referring to this idea:

Let up be positive:

5100 = 1/2at^2, where a = c -2g when c is constant acceleration and g is gravity.

c = 45.1 m/s/s when you solve for c?

8. Sep 24, 2011

### Ignea_unda

No, but you're getting closer. You know the acceleration due to gravity - it is a constant. You can look it up online or in your text book. And no one specified that the rate at which the rocket is accelerating upward is the same. So you have 2 accelerations for the first part "Ra" - the rocket acceleration due to it's engine, and 'g' - the negative acceleration due to gravity. In the second part, when the engine is not working any more, you have only 'g' acting. So if you sum those up (remember how long they both act), you set them equal to the height at the end, you will be able to solve for the rocket's acceleration 'Ra'.

9. Sep 24, 2011

### barthayn

I know I am getting close because I am only being off around 1000 m when I used a= c-2g and a=c-g. However, I do not know where to go from here. I am lost for ideas. I use A as a combination of them because I know you could (most likely anyways) express it as one acceleration for the total position of that rocket for 20 seconds. I don't see where one is combine to write it mathematically though. :(

EDIT: I also did this:

5100 = 1/2 at^2, where a = 16^2c - 4^2g
(10200 +16g)/16^2 = c
40.45 m/s/s/ = c.

When checking it by see if both adds up to 5100m in position I get the wrong answer. Is this closer than what I am before?

Last edited: Sep 24, 2011
10. Sep 24, 2011

### barthayn

Never mind. After about three days trying to figure this out. This question was very easy and I got the answer of 35.96 m/s/s. Thank you for your hint. It was very useful. I misread what you were trying to say at the time.