Rocket Acceleration, position and velocity

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SUMMARY

The acceleration of the rocket is defined by the equation a = Ct, where C equals 3 m/s³. To find the general position function x(t), one must integrate the acceleration function to derive velocity and subsequently integrate velocity to obtain position. At t = 5 seconds, with initial conditions x(0) = 0 and v(0) = 0, the calculations yield specific values for both position and velocity at that time.

PREREQUISITES
  • Understanding of calculus, specifically integration techniques.
  • Familiarity with kinematic equations in physics.
  • Knowledge of initial conditions in motion problems.
  • Basic grasp of the relationship between acceleration, velocity, and position.
NEXT STEPS
  • Study the process of integrating functions to find velocity from acceleration.
  • Learn how to apply initial conditions to solve differential equations in motion.
  • Explore the concept of concavity in graphs to interpret acceleration visually.
  • Investigate real-world applications of kinematic equations in rocket science.
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Students studying physics, particularly those focusing on kinematics and motion, as well as educators looking for examples of integrating acceleration to find position and velocity.

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Homework Statement



"The acceleration of a certain rocket is given by a=Ct, where C is a constant. (a) Find the general position function x(t). (B) Find the position and velocity at t=5 s if x= 0 and v= 0 at t=0 and C= 3 m/s^3"

so

a= Ct

t=5s
x=0

v=0
t=0
C= 3 m/s^3

Homework Equations



:confused:

The Attempt at a Solution


(a) Ill assume a "general position function" is just a small sketch of an "x vs t" Graph, since the rocket is accelerating the graph should just be a, concave curve up.

(B)I have no clue what exactly the question is asking for, with the given data.:frown:
 
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velocity is the integral of acceleration. position is the integral of velocity. Use that for the first part.
 
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?

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