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Find the minimum speed of a particle

  1. Oct 30, 2013 #1
    The position function of a particle is given by r(t)=⟨−5t2,−4t,t2+1t⟩.
    At what time is the speed minimum?

    I found the speed function by taking the magnitude of the velocity which is

    lv(t)l = sqrt(108t^2 +4t + 17)

    I then took the derivative of the speed to find the critical point. However, I'm always left with a negative time.

    d/dt lv(t)l = (104t + 2)/sqrt(104t^2 +4t+17)

    The only way to find the minimum is when the numerator is 0. But when I isolate t, it always equals a negative number.

    Is my approach correct?

    Any help would be appreciated. Thanks!
     
  2. jcsd
  3. Oct 30, 2013 #2

    Mark44

    Staff: Mentor

    Homework problems must be posted in the Homework & Coursework section, not in the technical math sections. I have moved this thread.
     
  4. Oct 30, 2013 #3

    Mark44

    Staff: Mentor

    That should be √(104t2 + 4t + 17).

    You can find the minimum value of |v(t)| by completing the square inside the radical. I get a negative time. All this denotes is that the speed is smallest at some time before time t = 0.
     
  5. Oct 30, 2013 #4
    But it doesn't make sense when t (time) is less than zero.
     
  6. Oct 31, 2013 #5

    HallsofIvy

    User Avatar
    Staff Emeritus
    Science Advisor

    Why not? t= 0 is just some arbitrarily chosen time. Why couldn't the particle be moving before that?
     
  7. Oct 31, 2013 #6

    pasmith

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    Homework Helper

    The other answer is that, if the domain of [itex]r(t)[/itex] is [itex][0,\infty)[/itex], then since [itex]\|v\|[/itex] is strictly increasing in [itex]t > 0[/itex] the particle attains its minimum speed at [itex]t = 0[/itex].
     
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