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

In summary: I then solved for t2t2 = 333 - (2 * vi2) + 1/2(a1(10.5^2) + 1/2(-9.8)(2^2))so t2 = 333 - 189 + 1/2(a1(10.5^2) + 1/2(-9.8)(2^2))so t2 = 333 - 251
  • #1
TheKracken5
27
7

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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  • #2
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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  • #3
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?
 
  • #4
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)
 

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

1. What is rocket acceleration and why is it important in space travel?

Rocket acceleration refers to the change in velocity of a rocket over time. It is important in space travel because it determines how quickly a rocket can reach its desired destination, and how much fuel is needed for the journey.

2. How is rocket acceleration calculated?

Rocket acceleration is calculated by dividing the change in velocity by the change in time. This can be represented by the equation a = Δv/Δt, where "a" is acceleration, "Δv" is the change in velocity, and "Δt" is the change in time.

3. What factors affect rocket acceleration?

The main factors that affect rocket acceleration are the thrust of the rocket, the mass of the rocket, and the resistance caused by air or gravity. The more thrust a rocket has, the faster it will accelerate. However, a heavier rocket or external forces like air resistance can slow down acceleration.

4. How does rocket acceleration change during a space mission?

During a space mission, rocket acceleration changes depending on the stage of the rocket and its distance from the Earth. At liftoff, the acceleration is at its highest as the rocket overcomes the force of gravity. As the rocket travels further from the Earth, acceleration decreases due to the decreasing force of gravity.

5. How do scientists use rocket acceleration to plan space missions?

Scientists use rocket acceleration to plan space missions by calculating the amount of thrust and fuel needed to reach a specific destination. They also consider the mass of the rocket and external factors that may affect acceleration. By understanding rocket acceleration, scientists can plan and execute efficient and successful space missions.

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