One more thing with motion problem

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Homework Help Overview

The discussion revolves around a motion problem involving the calculation of instantaneous velocity and acceleration from a position function. Participants are exploring the relationship between derivatives and motion concepts, particularly in the context of a specific function provided in the problem.

Discussion Character

  • Conceptual clarification, Mathematical reasoning

Approaches and Questions Raised

  • The original poster attempts to understand how to calculate the slope of the position-time curve for specific parts of the problem, expressing confusion about obtaining the instantaneous velocity. Other participants suggest using derivatives and clarify the relationship between position, velocity, and acceleration.

Discussion Status

Participants are actively discussing the methods to find instantaneous velocity and acceleration, with some providing guidance on using derivatives. There is a productive exchange of ideas, but no explicit consensus has been reached on the original poster's specific concerns.

Contextual Notes

The original poster references a previous problem and expresses uncertainty about the setup for calculating slopes at specific time points. The discussion includes the use of derivatives and the implications of the position function given in the problem.

frankfjf
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I'm referring to the same problem as before: https://www.physicsforums.com/showthread.php?t=107835

But now my question is how do I figure out parts c and e? My textbook says to just take the slope of the position-time curve. But if t = 1 in part c, how do I setup the slope? It'll just be something over 1, or that something. My problem is if I take the slope of the change in position over 1, I just get my answer to part a again, the average velocity or 71.3. I'm stuck. Please help?
 
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the instantaneous velocity function is given by the time-derivative of the position function.

do you have an expression for the velocity at some time t?
 
eep said:
the instantaneous velocity function is given by the time-derivative of the position function.

So does that mean that I take the derivative of x = 2.30^5 and plug in 1s and 2s, respectively for parts c and e?
 
yup! and the function for instantaneous acceleration is the time-derivative of the velocity function. One could also take the slope of the position function at t=1.0, since that's what the formula for the instantaneous velocity is giving you.
 
Thanks! So basically..

For the instantaneous velocity take the first derivative, and for the instantaneous acceleration take the second derivative? (Of x = 2.30t^5)
 
Correct. Of *any* position function.
 

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