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Kinematics and proportionality.

  • Thread starter raphile
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  • #1
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Hey, please take a look at this problem, it should be quite straight-forward but I want to know if I've reached the right conclusion.

Homework Statement



A particle moves in a straight line with displacement s from some fixed point on the line and velocity v.

If v is proportional to sqrt(s), find the behaviour of the acceleration with s; and if v=0 when t=0, find how v depends on time.

The Attempt at a Solution



I put v = k*sqrt(s) and used a = v*dv/ds = k*sqrt(s)*(1/2)k*s^(-1/2) = (1/2)k^2. So it seems that the s disappears, and therefore the acceleration is constant. Is it correct to say this?

For the second part, I put dv/dt=(1/2)k^2 (the acceleration found above) and solved to give v(t) = (1/2)(k^2)t + c, and c=0 because v=0 when t=0. Hence my conclusion is that v varies with t, i.e. v and t are proportional. Is this right? Thanks.
 

Answers and Replies

  • #2
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For the second part, I put dv/dt=(1/2)k^2 (the acceleration found above) and solved to give v(t) = (1/2)(k^2)t + c, and c=0 because v=0 when t=0. Hence my conclusion is that v varies with t, i.e. v and t are proportional. Is this right? Thanks.
If you let v=k[tex] \sqrt{s}[/tex], then the derivative with respect to s should be
[tex]= \frac{k}{2 \sqrt{s}}[/tex]
 
Last edited:
  • #3
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But in the part of my post that you quoted, I'm not using s at all. I'm just taking the acceleration that I found in the 1st part, (1/2)k^2, and calling it dv/dt, and then solving that to give v in terms of t. Is it not right to say v and t are proportional?
 
  • #4
Doc Al
Mentor
44,882
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I put v = k*sqrt(s) and used a = v*dv/ds = k*sqrt(s)*(1/2)k*s^(-1/2) = (1/2)k^2. So it seems that the s disappears, and therefore the acceleration is constant. Is it correct to say this?
Looks good to me.
For the second part, I put dv/dt=(1/2)k^2 (the acceleration found above) and solved to give v(t) = (1/2)(k^2)t + c, and c=0 because v=0 when t=0. Hence my conclusion is that v varies with t, i.e. v and t are proportional. Is this right? Thanks.
Again, looks good.
 

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