When does the particle reach its maximum x position?

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The discussion focuses on determining when a particle reaches its maximum position along the x-axis, described by the equation x = 3.5t^2 - t^3. To find the maximum x position, participants suggest taking the derivative of the position function, setting it to zero, and solving the resulting quadratic equation. The second derivative is also discussed to find acceleration at a specific time, with the correct expression being 2c - 6bt. Ultimately, the conversation emphasizes the importance of considering the coefficients in the equations and the necessity of selecting the positive time solution for 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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