Position, velocity, and acceleration

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SUMMARY

The discussion focuses on the motion of a particle along the x-axis described by the equation x(t) = 2.00 + 3.00 t – 1.00 t^2. At t = 3.00 s, the position of the particle is calculated using the given equation. The velocity is determined by taking the derivative of the position function, resulting in v(t) = 3.00 - 2.00 t, and evaluated at t = 3.00 s. The acceleration, derived from the velocity function, is constant at -2.00 m/s², indicating uniform acceleration throughout the motion.

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  • Understanding of calculus, specifically differentiation
  • Familiarity with kinematic equations
  • Knowledge of particle motion along a linear path
  • Ability to interpret graphs of functions and their slopes
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  • Learn how to derive position, velocity, and acceleration functions from kinematic equations
  • Study the concept of uniform acceleration in physics
  • Explore the application of derivatives in real-world motion problems
  • Investigate graphing techniques for visualizing motion functions
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Homework Statement



5. A particle moves along the x-axis, with x(t) = 2.00 + 3.00 t – 1.00 t^2.
SI units are used throughout.
(a) What is the position of the particle at t = 3.00 s?

(b) What is the velocity of the particle at t = 3.00 s?

(c) What is the acceleration of the particle at t = 3.00 s?


Homework Equations





The Attempt at a Solution



I solved letter a, but I have no idea how to complete letters b and c. Any help would be great.

Thanks.
 
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Take the derivative of the position function with respect to time and you get the velocity function of the object with respect to time. Plug in t =3s. Take the derivative of the velocity function with respect to time and you get the acceleration with respect to time which in this case you will find is a constant.

If you guys have not done calculus and are just doing this on a graphing calculator you can plot the position function and look a the slope at t = 3s and you will get the velocity at t = 3s. Now if your teacher gave you what those constants mean in front of the variable t, you can do the same thing with the velocity v. time and you will see a line. The slope of that line does not change and this tells you the acceleration is uniform through time.

Hope this helps some.
 
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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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