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

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The discussion focuses on calculating the acceleration of a rocket during its launch phase, specifically for the first 10.5 seconds of its ascent. The rocket has a mass of 50.0 kg and reaches an altitude of 333 m at 12.5 seconds, with gravity acting as a constant negative acceleration of 9.8 m/s². An initial attempt yielded an incorrect acceleration value of 3.29 m/s² due to not accounting for gravity properly. The correct approach involves using kinematic equations to relate the distances traveled during the acceleration phase and the subsequent free-fall phase, leading to a more accurate calculation of the rocket's acceleration. The discussion emphasizes the need for precise equation setup and consideration of all forces acting on the rocket.
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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