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how to find maximum speed???

sonic pwrsonic pwr Posts: 769Registered User Member
edited December 2005 in AP Tests Preparation
a particle moves on the x-axis in such a way that its position at time t is given by x=(2t-1)((t-1)^2)

i know that speed is absolute value of velocity and when velocity is 0 t= 2/3 and 1

so during the interval (2/3 , 1) when is the speed at a maximum??
Post edited by sonic pwr on

Replies to: how to find maximum speed???

  • aspariguyaspariguy Posts: 183Registered User Junior Member
    well you have the equation x=(2t-1)((t-1)^2)... so take the derivative of that to find eq for speed... first d second plus second d first...
    v(t) = (2t - 1)(2t - 2) + ((t - 1)^2)(2)
    v(t) = (4t^2 - 6t + 2) + (2t^2 - 4t +2)
    v(t) = 6t^2 -10t +4

    so then to find where speed is greatest, find the critical numbers (where deriv is 0 or undef). and i know i'm working with the velocity eq... so we'll count min's as well as max's to find "max speed"

    v'(t) = 12t - 10
    0 = 12t - 10
    t = 5/6

    so plug in 5/6 to velocity

    v(5/6) = 6*(5/6)^2 - 10(5/6) - 4
    ..........= 25/6 - 50/6 + 24/6
    ..........= -1/6
    take absolute value, and the max speed is 1/6 units/sec

    i guess you didn't need to know what the max speed was, but i did it anyway. but the t value for the max is 5/6.

    -will
  • sonic pwrsonic pwr Posts: 769Registered User Member
    also just to make sure, on what interval is the particle moving to the left? i got (2/3,1) is that right?
  • BCCBCC Posts: 157Registered User Junior Member
    Is that an open or closed interval? If it's a close interval you must check the endpoints for speed.
    abs(v(2/3)) = abs(v(1)) < abs(v(5/6))
    K, answer for max speed still occurs at time = 5/6

    The particle moves to the left when velocity is negative, so the entire interval, yes.
  • sonic pwrsonic pwr Posts: 769Registered User Member
    entire interval meaning [2/3,1] or (2/3,1)
  • BCCBCC Posts: 157Registered User Junior Member
    Actually, we use [2/3,1] to mean 2/3<=t<=1 and (2/3,1) to mean 2/3<t<1

    So, you see, with the first notation you actually have extremas at the endpoints. Hence why you needed to check if speed was greatest there.
  • aspariguyaspariguy Posts: 183Registered User Junior Member
    you don't need to check the endpoints in this one because he said straight out that the speed at the endpoints IS zero.
  • BCCBCC Posts: 157Registered User Junior Member
    Ouch, totally missed that. =(
    But c'mon, who is never been guilty of not reading a question completely? =P
  • sonic pwrsonic pwr Posts: 769Registered User Member
    so is it [2/3,1] or (2/3,1)? which one is it? the question was asking, during what interval of time is the particle moving to the left?
  • BCCBCC Posts: 157Registered User Junior Member
    At t = 2/3 and t = 1, since velocity is zero, you are in the process of changing directions (you are not moving to the left).
    So you move to the left at 2/3<t<1 or (2/3,1).
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