What is the velocity and acceleration of a particle at t = 15.0 s?

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Homework Help Overview

The problem involves determining the velocity and acceleration of a particle based on its position function, which is expressed as a quadratic equation in terms of time. The constants in the equation represent specific values for position, velocity, and acceleration.

Discussion Character

  • Exploratory, Mathematical reasoning, Assumption checking

Approaches and Questions Raised

  • Participants discuss the differentiation of the position function to find velocity and acceleration. There are questions about the interpretation of the variables in the equation and the nature of the constants involved.

Discussion Status

Some participants have provided guidance on how to differentiate the position function and have confirmed the relationships between position, velocity, and acceleration. There is ongoing clarification regarding the constants and the differentiation process, with some participants checking their understanding of derivatives.

Contextual Notes

One participant expresses confusion about the equation and the differentiation process, indicating a potential gap in understanding the underlying concepts. There is mention of a LaTeX tutorial for formatting equations, which suggests that participants are navigating both mathematical and technical aspects of the discussion.

Zhalfirin88
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New to PF, so here it goes. Oh just a question, did this site use to have a dark skin? I can't remember if that was this or something else.

Homework Statement


The position of a particle as a function of time (in s) is given by C1 + C2t + C3t2. Let C1 = 11.0 m, C2 = 9.5 m/s and C3 = -0.49 m/s2. What is the velocity of the particle at time t = 15.0 s? And what is the particle's acceleration at time t = 15.0 s?

Homework Equations


Velocity is the derivative of position.(How do you make it look all nice?)
Acceleration is the 2nd derivative of position.

The Attempt at a Solution



Where do I start? I don't even understand the equation that's given. Judging from the question (it's online) it looks like those t are variables and not subscripts.
 
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Zhalfirin88 said:
New to PF, so here it goes. Oh just a question, did this site use to have a dark skin? I can't remember if that was this or something else.

Homework Statement


The position of a particle as a function of time (in s) is given by C1 + C2t + C3t2. Let C1 = 11.0 m, C2 = 9.5 m/s and C3 = -0.49 m/s2. What is the velocity of the particle at time t = 15.0 s? And what is the particle's acceleration at time t = 15.0 s?


Homework Equations


Velocity is the derivative of position.(How do you make it look all nice?)


The Attempt at a Solution



Where do I start? I don't even understand the equation that's given. Judging from the question (it's online) it looks like those t are variables and not subscripts.

Welcome to the PF. The skins have changed several times, so it's possible that a previous skin was dark.

The "t" in the equation is time, so you would just plug in the time to get the position as a funtion of time. You are correct about the velocity being the derivative of the position, and then the acceleration is the derivative of the velocity. Using LaTex, it looks like this:

[tex]x(t) = C_1 + C_2 t + C_3 t^2[/tex]

[tex]v(t) = \frac{dx(t)}{dt} = \frac{d(C_1 + C_2 t + C_3 t^2)}{dt}[/tex]

[tex]a(t) = \frac{dv(t)}{dt}[/tex]

So just do the differentiations, and plug in the respective times to get the answers.
 
Just to check my derivatives since I haven't done them in months, I'm assuming that the C's aren't constants, right?

I don't know the syntax for LaTeX so, the derivative of postion would be:

C2 + 2C3t

2nd derivative would be:

2C3
 
Correct. To see how the LaTex is formed, you can just QUOTE my post, to see the tags that are embedded in it. Also, there is a stickie thread with a LaTex tutorial at the top of the Learning Materials forum:

https://www.physicsforums.com/forumdisplay.php?f=151

.
 
I don't understand why it's saying my answer for the acceleration is wrong. My velocity answer was correct though.

2C3 = 2(-0.49) = -0.98 m/s2

Thanks for the sticky, I'll check it out when I finish my homework :P

Ah never mind, forgot to type in the negative sign -.-
 

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