When does the particle reach its maximum x position?

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Homework Help Overview

The problem involves determining when a particle reaches its maximum position along the x-axis, based on the equation x = ct² - bt³, with specific values for c and b. Additionally, there is a question regarding the particle's acceleration at a given time.

Discussion Character

  • Exploratory, Conceptual clarification, Mathematical reasoning, Assumption checking

Approaches and Questions Raised

  • Participants discuss the need to find the derivative of the position function to locate the maximum position. There are attempts to solve for time by setting the first derivative to zero and identifying the nature of the solutions. Questions arise about the implications of coefficients being zero and how to interpret the results.

Discussion Status

Some participants have provided guidance on taking derivatives and setting them to zero to find critical points. There is ongoing exploration of the implications of the quadratic equation's solutions, with some uncertainty about the validity of the results and the conditions under which they apply.

Contextual Notes

Participants are questioning the assumptions regarding the coefficients c and b, particularly whether they can be zero. There is also a discussion about the physical interpretation of negative time and initial conditions affecting the maximum position.

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Homework Statement



The position of a particle moving along the x-axis depends on the time according to the equation x = ct2 - bt3, where x is in meters and t in seconds.

For the following, let the numerical values of c and b be 3.5 and 1.0 respectively.

(b) At what time does the particle reach its maximum positive x position?

(f) What is its acceleration at at t = 1.0 s?

Homework Equations





The Attempt at a Solution



For (b) I tried plugging in numbers starting at 1 until the the result started to decrease. I am not sure if I must take the derivative first though.

For (f) I am assuming you take the second derivative of the equation. Is the derivative not 2-6bt ? Answer I got : -4
 
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Well what does dx/dt represent? The rate of change.

now I can't tell, is it sposed to be x=c*t^2-b*t^3? I'm assuming or else it's a terribly uninteresting problem

So for b

dx/dt=2*c*t-3*b*t^2

you're looking for a maximum, so you set dx/dt=0 and solve for t. It's a quadratic equation so there are two solutions

How do you decide which one's right?

You had the right idea for f, but the second derivative is 2c-6bt=a
 
(Use the symbol ^ for power.)

x = ct^2 - bt^3.

b. At the extreme +ve posn the v would be 0, so, yes, you should find dx/dt.

f. 2c-6bt. Now find answer.
 
Thank you, I got (f) right. But for (b) I have another question. For the quadratic equation does c = 0? To decide which one is right, you pick the positive one because time cannot be negative. For my answer I am getting 0. This doesn't sound right.
 
it's possible if the initial velocity is negative and it only gets more negative, then the initial position should be the biggest it ever gets
 
suxatphysix said:
But for (b) I have another question. For the quadratic equation does c = 0? To decide which one is right, you pick the positive one because time cannot be negative. For my answer I am getting 0. This doesn't sound right.

For elementary problems like these, unless otherwise mentioned, the co-efficients will not be equal to zero. But it's shows your interest that you have asked, and blochwave has given you the answer. You should then also ask what if c=0, both=0.

Assuming non-zero b and c, have you solved it?
 

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