Displacement of zigzaging dust particle

AI Thread Summary
A dust particle bombarded by air molecules travels in a zigzag path at constant speed. In part (a), the distance traveled in time t is straightforwardly calculated as vt. Part (b) involves determining the displacement vector after N steps, where each step is random; the average magnitude of the resultant vector is derived using vector addition principles. The discussion highlights confusion around using the dot product and averaging angles, ultimately leading to the conclusion that the displacement is d√N. The participants clarify their understanding of the average value of cosine in this context, leading to a resolution of the problem.
Pseudopro
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Homework Statement


A particle of dust is bombarded by air molecules and follows a zigzag path at constant speed v.
(a)Assuming each step has a length d, find the distance traveled by the dust particle in time t.
(b)What is the length of the displacement vector after N steps where N is large? Assume that each step is taken in a random direction on the plane. (This problem assumes you are familiar with the scalar product of two vectors.

Homework Equations


v2=(v12+v22+...+vN2)/N

The Attempt at a Solution


The answer to part (a) seemed quite obvious: vt.
However, part (b) isn't so easy. I've been able to get the answer d\sqrt{N} by doing vt=t x sqrt[(v12+v22+...+vN2)/N] but this is getting the answer for the sake of it. My method doesn't actually make any sense because it can cancel out to nothing. Please help!
(I believe this is called Brownian motion?) (the thing I don't get is how do I use dot product in this instance?)
 
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Hi Pseudopro! :smile:

Hint: if you add two vectors whose lengths are fixed as a and b, but whose directions are random,

what is the average magnitude of the sum of those vectors? :wink:
 
(a+b)/2?
 
uhh? :confused:

call the angle θ, and work out the magnitude! :rolleyes:

(and then average over θ)
 
Sorry, I don't think I'm doing this properly (or I don't have the knowledge).

Are you looking for something like this? \sqrt{a^{2}+b^{2}-2abcos\theta}
 
That's it! :smile:

ok, since θ can be anything, the average value of that is (roughly) … ? :wink:

(and then try the same thing for n vectors)
 
average value of cos would be 0.5 I think - so it would end up as d... (wait does the assumption from part (a) still hold?) I don't see how I can get sqrt N... if I put d back into the equation, I just get exactly the same thing back again...
 
Pseudopro said:
average value of cos would be 0.5 I think …

uhh? :confused: average value of cos is zero
 
oh yeah. my bad...
 
  • #10
Thanks a lot for your guidance. :smile: I've got it now :approve:
 
  • #11
I am also doing this problem and understand the hints above, but I don't understand how the answer is d\sqrt{N}. Can someone help?? :rolleyes:
 
  • #12
I am also doing this problem and understand the hints above, but I don't understand how the answer is d√N. Can someone help?? :rolleyes:
 
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