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Derivative (problem solving)

  1. Dec 11, 2009 #1
    Let s = s(t) be the position function of a particle moving in a straight line.
    Suppose that the position of the particle is given by the formula
    s (t) = t^2 e^-t ; t >= 0
    where t is measured in seconds and s in meters.

    (i) Find the velocity of the particle at time t.
    * should i differentiate s(t) to get the velocity?

    (ii) When is the particle at rest?
    * when velocity = 0.. am i right?

    (iii) Find the total distance traveled by the particle during the first two seconds.
    *dont have any idea..
     
  2. jcsd
  3. Dec 11, 2009 #2

    Dick

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    Science Advisor
    Homework Helper

    (i) yes. (ii) yes. (iii) you might have a more informed opinion about this after you do (i) and (ii). Get started.
     
  4. Dec 12, 2009 #3
    (i) Find the velocity of the particle at time t.
    * should i differentiate s(t) to get the velocity?

    answer--> (2t e^-t) - (t^2 e^-t)

    (ii) When is the particle at rest?
    * when velocity = 0.. am i right?

    answer-->
    (2t e^-t) - (t^2 e^-t) = 0
    (2t e^-t) = (t^2 e^-t)
    (2e^-t) = (t e^-t)
    *how to solve t?

    (iii) Find the total distance traveled by the particle during the first two seconds.
    *dont have any idea..

    at t = 2
    s (t) = t^2 e^-t
    s (2) = 2^2 e^-2
    .......= 0.54 meters

    is my answer for (i) and (iii) is correct?
    how am i going to solve (ii)?
     
  5. Dec 12, 2009 #4
    Are you just trying to solve (ii) for t?
     
  6. Dec 12, 2009 #5
    yes. because i need the value of 't' when the particle at rest..
     
  7. Dec 12, 2009 #6
    do you see anything you can factor out from both sides?
     
  8. Dec 12, 2009 #7
    (2t e^-t) - (t^2 e^-t) = 0
    (2t e^-t) = (t^2 e^-t)
    (2e^-t) = (t e^-t)
    e^-t(2) = e^-t (t)
    (e^-t)/(e^-t) 2 = t

    *(e^-t)/(e^-t) = 1?
     
  9. Dec 12, 2009 #8

    Mark44

    Staff: Mentor

    You should write an equation; namely v(t) = 2te^(-t) - t^2*e^(-t)
    Yes.
    I'm going to cut in here because the rest of your work doesn't help you get where you need to go. It's not wrong, but it isn't helpful either.
    s'(t) = v(t) = 2t e^(-t) - t^2 e^(-t) = 0
    v(t) = 0 ==> 2t e^(-t) - t^2 e^(-t) = 0 = 0 ==> (2t - t^2)e^(-t) = 0
    e^(-t) is never 0. When is 2t - t^2 = 0? Those are the times when v(t) = 0.

    Integrate the velocity between t = 0 and t = 2.
     
  10. Dec 12, 2009 #9
    for (iii) 4e^-2
    am i got it right?
     
    Last edited: Dec 12, 2009
  11. Dec 12, 2009 #10

    HallsofIvy

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    Staff Emeritus
    Science Advisor

    Yes, of course. You were told in the problem itself that s(t)= t2e-t. The answere to (iii) is just s(2).
     
  12. Dec 12, 2009 #11
    Thank u for the confirmation of my answer.. =)
     
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