Acceleration in terms of displacement

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SUMMARY

The discussion focuses on solving a physics problem involving acceleration and displacement, specifically finding velocity at x=3 and the time taken to travel from x=1 to x=3. The original differential equation is given as dv/dt = 2x, which transforms to dv/dx * v = 2x. By applying the initial condition of v=0 at x=1, the constant u is determined to be -2, leading to the equation v^2 = 2(x^2 - 1). For the time calculation, the relationship v = dx/dt is utilized, resulting in the integral of dx/sqrt(x^2 - 1) from 1 to 3 to find t.

PREREQUISITES
  • Understanding of calculus, specifically integration and differentiation
  • Familiarity with differential equations
  • Knowledge of kinematics, particularly the relationships between velocity, acceleration, and displacement
  • Ability to manipulate and solve algebraic equations
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  • Study the application of the chain rule in calculus
  • Learn about solving first-order differential equations
  • Explore kinematic equations in physics for motion analysis
  • Practice integration techniques, particularly with square roots in the denominator
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This discussion is beneficial for physics students, particularly those studying mechanics, as well as educators and tutors looking to enhance their understanding of calculus applications in kinematics.

ykeyani
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Homework Statement



a)find the velocity when x = 3, b)find the time taken to travel between x=1 to x=3.

when x = 1, v = 0

Homework Equations



a=2x

The Attempt at a Solution



a)using the chain rule I got v = 2x^2 + u (u as a constant), but I think that's wrong.
i tried using dx/dt = dv/dt * dx/dv
b)I integrated that with respect to time giving 3/16, but I think the solution there is wrong too.

I'm not really sure how to tackle this type of problem, and I'm pretty sure I'm missing something here, my book hasn't shown me any examples so far relating to this so I hope someone can help me, I'm a bit rubbish at calculus anyway and there's no answer in the back to guide me.
 
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It was a good start, but the result is not correct. The original differential equation is

[tex]\frac{dv}{dt}=2x[/tex]

which can be transformed to

[tex]\frac{dv}{dx}v=2x[/tex]

but the integration results in

[tex]v^2=2x^2+u[/tex].

Use the condition x=1, v= 0 to get u. (u=-2, so v^2=2(x^2-1). Then you can find v when x=3.

For question b, you need the x(t) function. v=dx/dt.

[tex]\frac{dx}{dt}=\sqrt{2(x^2-1)}\rightarrow \frac{1}{\sqrt{2}}\int_1^3{\frac{dx}{\sqrt{x^2-1}} = t[/tex].


ehild
 
thanks that was a big help for me, so I really appreciate it, I worked through it again and I understand it now.
 

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