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The velocity of a particle at the origin.

  1. Sep 25, 2012 #1
    1. The problem statement, all variables and given/known data
    A particle moves according to the position function x(t)=ct2+bt where c=3m/s2 and b=-7m/s. Find the velocity at the origin.


    2. The attempt at a solution
    I tried just taking the derivative and setting t=0 to the equation as so:
    vx(t)=6t-7=6(0)-7=-7 m/s
    Although, when I put the answer into WebAssign it says I'm incorrect. I'm not looking for a definite answer since this is my homework but would just like to understand what I am doing wrong.
    1. The problem statement, all variables and given/known data



    2. Relevant equations



    3. The attempt at a solution
     
  2. jcsd
  3. Sep 25, 2012 #2
    Maybe they're looking for the velocity at the other time the particle is at the origin?
     
  4. Sep 25, 2012 #3
    I'm under the impression that the origin means they want me to solve it at t=0. What other origin could they be asking for?
     
  5. Sep 25, 2012 #4
    I guess they mean the origin of the coordinate system, i.e. x=0. You usually say t=0 rather than the slightly ominous "the origin of time".
     
  6. Sep 25, 2012 #5
    Yeah I just tried now to solve it at x=0 where I find that,
    x(t)= 3t2-7t
    0=3t2-7t+0constant
    Then I treated it like a quadratic:
    t=7+√[72-4(3)(9)] = 7/3s
    2(3)
    Then I put 7/3s into the formula...and...I got it right, thanks so much.



    Answer ends like so,
    vx=6(7/3)-7= 7m/s
     
  7. Sep 25, 2012 #6
    There is an easier way to solve it, just factor out one of the t's:
    x=ct^2+bt = t(ct+b),
    for which x=0 when t=0 or t=-b/c. Both solutions are of course valid (based on the problem text), even if the homework website doesn't think so...
     
  8. Sep 25, 2012 #7
    x(t) = 0 when t=2.3333333333
     
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