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

Click For Summary
SUMMARY

The correct acceleration of a 50.0 kg rocket during its launch phase, which lasts for 10.5 seconds, is determined to be approximately 3.29 m/s². However, this value is incorrect as it does not account for the negative acceleration due to gravity (g = 9.80 m/s²). The accurate approach involves using the equations of motion: d = v₀t + 1/2at² and v = v₀ + at, while considering the total distance traveled (333 m) and the effects of gravity during the rocket's ascent and subsequent free fall.

PREREQUISITES
  • Understanding of kinematic equations: d = v₀t + 1/2at² and v = v₀ + at
  • Basic knowledge of physics concepts such as acceleration and gravity
  • Ability to manipulate algebraic equations to solve for unknowns
  • Familiarity with concepts of projectile motion and free fall
NEXT STEPS
  • Review the derivation and application of kinematic equations in physics
  • Learn how to incorporate forces, such as gravity, into motion equations
  • Explore differential equations (DEQ) and linear algebra (LA) methods for solving motion problems
  • Practice solving similar physics problems involving projectile motion and acceleration
USEFUL FOR

Students studying physics, particularly those focusing on mechanics, as well as educators looking for effective teaching methods for kinematics and motion analysis.

TheKracken5
Messages
27
Reaction score
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!
 
Physics news on Phys.org
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.
 
  • Like
Likes PeroK
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)
 
The book claims the answer is that all the magnitudes are the same because "the gravitational force on the penguin is the same". I'm having trouble understanding this. I thought the buoyant force was equal to the weight of the fluid displaced. Weight depends on mass which depends on density. Therefore, due to the differing densities the buoyant force will be different in each case? Is this incorrect?

Similar threads

  • · Replies 3 ·
Replies
3
Views
2K
  • · Replies 1 ·
Replies
1
Views
2K
  • · Replies 3 ·
Replies
3
Views
1K
Replies
8
Views
4K
  • · Replies 11 ·
Replies
11
Views
2K
  • · Replies 16 ·
Replies
16
Views
9K
  • · Replies 3 ·
Replies
3
Views
1K
Replies
14
Views
5K
Replies
9
Views
25K
  • · Replies 10 ·
Replies
10
Views
9K