Mechanics- general motion in a straight line

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SUMMARY

The discussion centers on calculating the derivative of the motion equation \(s = 4t^2 - \sqrt{2t^3}\). Participants confirm that knowledge of calculus, specifically derivatives, is essential for solving this problem. The derivative is calculated as \(ds/dt = 8t - \frac{3\sqrt{2t}}{2}\), leading to a quadratic equation that simplifies to \(t = 2s\). This indicates a direct relationship between time and displacement in the context of motion in a straight line.

PREREQUISITES
  • Understanding of calculus, specifically derivatives
  • Familiarity with motion equations in physics
  • Ability to manipulate algebraic expressions
  • Knowledge of quadratic equations
NEXT STEPS
  • Study the principles of calculus derivatives in detail
  • Explore motion equations in physics, focusing on their applications
  • Practice solving quadratic equations and their implications in motion
  • Learn about the relationship between displacement, velocity, and time
USEFUL FOR

Students studying physics and calculus, educators teaching motion concepts, and anyone interested in the mathematical analysis of motion in a straight line.

Shah 72
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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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