Particle Movement Homework: Find x,y,z Velocity & Acceleration

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Homework Help Overview

The problem involves determining the velocity and acceleration components of a particle moving in three-dimensional space, described by a position vector as a function of time. The components are expressed in terms of the natural logarithm and exponential functions.

Discussion Character

  • Exploratory, Mathematical reasoning, Assumption checking

Approaches and Questions Raised

  • Participants discuss taking derivatives of the position components to find velocity and acceleration. There are questions about the correct application of derivatives, particularly with fractions and the process of differentiation.

Discussion Status

Some participants have provided guidance on how to differentiate the components, while others are clarifying their understanding of the steps involved. There is an acknowledgment of confusion regarding the identification of components and the differentiation process, but productive dialogue is ongoing.

Contextual Notes

There appears to be some misunderstanding regarding the identification of the velocity and acceleration components, leading to a need for clarification on the derivatives being calculated. Participants are also navigating the challenge of working with fractional expressions in their calculations.

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Homework Statement


A particle moves so that its position (in meters) as a function of time (in seconds) is R= ln(1+t)i+ (et-1)j + tk . Write expressions for (a) x-, (b) y- and (c) z-components of its velocity as well as the (d) x-, (e) y- and (f) z-components of its acceleration as functions of time.


Homework Equations





The Attempt at a Solution



x-component= 1/1+t
y-component= et
z-component= 1

How do I find their components of acceleration as a function of time??


 
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Take the derivative of a particular position component with respect to time twice. For example

[tex]v_{x}=\frac{dx}{dt}[/tex]

[tex]a_{x}=\frac{dv_{x}}{dt}[/tex]
 
Ok so for the x-component 1/1+t...the derivative of that would be ...?? I am not good with fractions..Once I have that, I put it over 2t?
 
Write the fraction as (1 + t)-1, then take its derivative the usual way.
 
so...-(1+t)^-2?

then the next one will just say e^t?
And the final one is 0??

I found that derivative, but what do you mean with the twice time? Or is my work done?
 
You have found that vx=-(1+t)-2. To find the acceleration, you need to take the derivative one more time. So, to get the acceleration, you need to take the derivative two times, one to get the velocity and one more to get the acceleration. Once you get that, your work is not done yet. You need to repeat for the y and z components of the position.
 
Ok I put it for the x ..2(1+t)^-3...and it was wrong..That is the second derivative though?
 
That is the second derivative of x, i.e. the x-component of the acceleration. What exactly do you mean "put it for the x"? What was the question that you used it as an answer to?
 
I used that answer for part D, the x-component of acceleration.
 
  • #10
I am sorry I misled you. :redface: I looked at what you had written down

x-component = 1/(1+t)

and I assumed that this was the x-component of the position. It is actually the velocity component. So you need to take the derivative of that to get the acceleration, which in this case gives you ax=-(1+t)-2. I apologize for the confusion it might have caused you. Your work was actually done when you asked if it was.
 
  • #11
No problem! I put the first derivatives in as the answers, and its right! Thanks for the help, and don't worry about the confusion, its ok!
 

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