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A Particle's Velocity

  1. Jan 24, 2010 #1
    1. The problem statement, all variables and given/known data

    A particle's velocity is described by the function [tex]v_x=kt^2 m/s[/tex], where k is a constant and t is in s. The particle's position at [tex]t_0[/tex]=0 is [tex]x_0[/tex] = -6.90 . At [tex]t_1[/tex]= 1.00, the particle is at [tex]x_0[/tex]= 8.70 .


    Determine the units of k in terms of m and s.
    2. Relevant equations

    None

    3. The attempt at a solution

    So I really have no idea what this question is asking. I tried integrating the velocity equation then solving for c and eventually k getting a value of 46.8. It's online homework and it wouldn't take that, even if I added m/s to the end.

    I am pretty sure what I tried isn't remotely close to what I am supposed to be doing. Any ideas?
     
  2. jcsd
  3. Jan 24, 2010 #2

    ideasrule

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    Aren't you supposed to be finding the units instead of calculating any numerical values? What unit do you multiply by s^2, the unit of t^2, to get m/s, the unit of v?
     
  4. Jan 24, 2010 #3
    Yeah, you are right. Although I am almost more confused now. I seriously have no idea what I am supposed to do or what the answer is supposed to look like. It's a first for me so I am completely dumbfounded. :rofl:

    Am I just supposed to get rid of the t^2 so k is alone with m/s?

    For some reason I am always stumped on these extremely easy problems. I don't get them. It's pretty sad.
     
  5. Jan 24, 2010 #4

    ideasrule

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    Here's an example: if k has the unit "s", kt^2 would have the unit s^3 (s * s^2 = s^3). If k has the unit "m", kt^2 would have the unit ms^2 (m * s^2 = ms^3). What unit, multiplied by s^2, gives m/s? You can treat the units the same way you treat algebraic variables.
     
  6. Jan 24, 2010 #5
    m/s^3?

    I think why I am so confused it because it already has m/s in the first part. Does that even matter?
     
  7. Jan 24, 2010 #6

    ideasrule

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    That's just to tell you that v is in m/s. It really shouldn't be written like that; I can see why it would confuse the hell out of a lot of students.

    But yes, m/s^3 is right.
     
  8. Jan 24, 2010 #7
    Thank you, thank you, thank you. There is no way I would have figured that out on my own.

    That original m/s was definately throwing me off. Since it almost seemed like it was already in m/s terms...almost.

    Thanks again.
     
  9. Jan 24, 2010 #8
    On a side note the next question was to figure out the value for k. Which I had correct from my original incorrect answer. So that makes me feel a bit better about it.:tongue2:
     
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