What is the Correct Acceleration of the Rocket During Its Launch Phase?

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Homework Help Overview

The discussion revolves around the acceleration of a rocket during its launch phase, specifically focusing on the first 10.5 seconds of its ascent. The problem involves kinematic equations and the effects of gravity on the rocket's motion.

Discussion Character

  • Exploratory, Mathematical reasoning, Assumption checking

Approaches and Questions Raised

  • Participants discuss modeling the problem using kinematic equations, attempting to separate the motion into segments of acceleration and free fall. There is a focus on determining the correct acceleration during the initial phase and the total distance traveled.

Discussion Status

Some participants have provided insights into the equations used and have questioned the assumptions made regarding the effects of gravity. There is an ongoing exploration of the calculations and the need for clarity in the approach taken.

Contextual Notes

Participants note the importance of considering gravity as a negative acceleration and the implications of this on the overall calculations. There is also mention of the desire to explore alternative methods such as differential equations or linear algebra for solving the problem.

TheKracken5
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Homework Statement


A 50.0 kg rocket is launched straight up (we’ll call this the y direction). Its motor produces constant acceleration for 10.5 seconds and stops. At the time of 12.5 seconds the altitude of this rocket is 333 m. (ignore air resistance and take g=9.80m/s^2)
a. What is the rockets acceleration during the first 10.5 seconds?

Homework Equations


d = v0t +1/2at^2
v = v0 +at

The Attempt at a Solution


So, I keep trying to model this question with 2 equations. And setting d1 = 333 - d2 (d1 being distance traveled by rocket while rocket was accelerating and d2 being the remaining distance out of the 333m total traveled) and using this method I try to find a1 ( the acceleration of the rocket during the first 10.5 seconds) and using a2 = -9.8m/s^s for gravity.

After running through this, I got the acceleration to be 3.29m/s^2 for the rocket, but when I use this to see the total distance traveled in 10.5s + the distances traveled the last 2 seconds, I get something like 240m total, which is obviously wrong.What is the best way to approach this problem? Also if there is a way to solve this problem using DEQ's or LA to make it easier, that would be nice to know as well!
 
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It would be helpful if you showed the equations that you got from your solution description. Paragraphs of description are vague compared to showing the precise equations and calculations.

I don't see anything wrong in your description, but that still leaves a lot to guess at.
 
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TheKracken5 said:

Homework Statement


A 50.0 kg rocket is launched straight up (we’ll call this the y direction). Its motor produces constant acceleration for 10.5 seconds and stops. At the time of 12.5 seconds the altitude of this rocket is 333 m. (ignore air resistance and take g=9.80m/s^2)
a. What is the rockets acceleration during the first 10.5 seconds?

Homework Equations


d = v0t +1/2at^2
v = v0 +at

The Attempt at a Solution


So, I keep trying to model this question with 2 equations. And setting d1 = 333 - d2 (d1 being distance traveled by rocket while rocket was accelerating and d2 being the remaining distance out of the 333m total traveled) and using this method I try to find a1 ( the acceleration of the rocket during the first 10.5 seconds) and using a2 = -9.8m/s^s for gravity.

After running through this, I got the acceleration to be 3.29m/s^2 for the rocket, but when I use this to see the total distance traveled in 10.5s + the distances traveled the last 2 seconds, I get something like 240m total, which is obviously wrong.What is the best way to approach this problem? Also if there is a way to solve this problem using DEQ's or LA to make it easier, that would be nice to know as well!

As you know, ##3.29ms^{-2}## is not correct. How did you get that?
 
PeroK said:
As you know, ##3.29ms^{-2}## is not correct. How did you get that?

As I typed this all out I found my error, I didnt factor in gravity as a negative acceleration. Since I typed out most of my work, I guess I will post it anyway. Now onto further parts of this problem and will post if I have further issues.

d1 = vi1 * t1 + 1/2(a1) *( t1)^2
d2 = vi2 * t2 + 1/2(a2) * (t2)^2

since d2 = 333-d1 I plug this into the second equation above and get

333 - (vi1 * t1 + 1/2(a1) *( t1)^2 ) = vi2 * t2 + 1/2(a2) * (t2)^2

we also know that vi2 = a1*t and we know vi1 = 0
I moved stuff over and plug stuff in

333 = (2 * vi2) + 1/2(a1(10.5^2) + 1/2(-9.8)(2^2)
 

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