Acceleration in terms of displacement

AI Thread Summary
The discussion focuses on solving a physics problem involving acceleration, displacement, and velocity. The original differential equation is given as dv/dt = 2x, which is transformed to dv/dx * v = 2x for integration. The correct approach leads to the equation v^2 = 2(x^2 - 1), allowing for the determination of velocity at x = 3. For the time taken to travel from x = 1 to x = 3, the integration of dx/dt = √(2(x^2 - 1)) is necessary. The participant expresses gratitude for the clarification, indicating a better understanding of the problem.
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

\frac{dv}{dt}=2x

which can be transformed to

\frac{dv}{dx}v=2x

but the integration results in

v^2=2x^2+u.

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.

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


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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