Mechanics- general motion in a straight line

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Discussion Overview

The discussion centers around calculating motion in a straight line using a given position function, specifically exploring the application of calculus, particularly derivatives, to find velocity and solve for time.

Discussion Character

  • Homework-related

Main Points Raised

  • One participant expresses uncertainty about how to calculate the motion described by the position function.
  • Another participant provides the position function $s = 4t^2 - \sqrt{2t^3}$ and asks if calculus, specifically derivatives, has been learned.
  • A participant calculates the derivative of the position function, stating that $ds/dt = 8t - \frac{3\sqrt{2t}}{2}$.
  • Another participant attempts to set the derivative equal to a velocity value (13) and formulates an equation to solve for time.
  • A participant concludes they derived a quadratic equation and found a relationship $t = 2s$, expressing gratitude for the assistance.

Areas of Agreement / Disagreement

The discussion includes multiple contributions with varying approaches to the problem, but no consensus is reached on the final solution or the correctness of the derived relationships.

Contextual Notes

Participants do not clarify assumptions regarding the values used in their calculations, nor do they resolve the mathematical steps leading to the final expressions.

Shah 72
MHB
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I don't know how to calculate this. Pls help
 
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$s = 4t^2 - \sqrt{2t^3}$

have you learned any calculus, specifically derivatives?
 
skeeter said:
$s = 4t^2 - \sqrt{2t^3}$

have you learned any calculus, specifically derivatives?
Yeah so ds/dt=8t-3sqroot(2t)/2
 
skeeter said:
$s = 4t^2 - \sqrt{2t^3}$

have you learned any calculus, specifically derivatives?
I have tried doing this so ds/dt=v
13= 8t-3sqroot (2t)/2
 
skeeter said:
$s = 4t^2 - \sqrt{2t^3}$

have you learned any calculus, specifically derivatives?
So I get quadratic equation. I got t=2s. Thank you very much!
 

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