Sunday, April 19, 2009

A particle moves in a straight line and at point P, it's velocity is given as v = 7t^2 - 5t +3. The particle comes to rest at point Q.

1. What is the acceleration at Q if it arrives at Q when t=7?
2. How far does the particle travel in t=1 to t=4?
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6 comments:

  1. crazykidApril 19, 2009 at 12:53 PM

    well i can work the acceleration of the particle
    where v = 7t^2 -5t +3

    dv/dt= 14t - 5

    if the particle comes to rest at point Q, then there shouldn't b any acceleration at that point.

    therefore acceleration should be = 0

    ok i really don't know.....can some1 help me

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  2. Swanky ThingApril 19, 2009 at 1:11 PM

    I think your on the right track with the acceleration part But am also lost with the acceleration at point Q.

    Do you think that it suppose to be dv/dQ we suppose to be finding, using a substitution....

    Am not sure here guys, but help me out....

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  3. Swanky ThingApril 19, 2009 at 1:13 PM

    For part 2, i know that when you find dv/dt all you have to do now is to substitute the value of t=1 and t=4 in it to find how far the perticle travelled..

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  4. johnsmithApril 20, 2009 at 12:28 PM

    i just think it is a trick question becasue at rest the acceleration will be 0 so i agree with crazykid

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  5. Fariel MohanApril 22, 2009 at 2:30 PM

    Post corrected.

    ReplyDelete
  6. turboApril 28, 2009 at 1:17 AM

    Swanky Thing, u'r wrong.To find dist. you need to intergrate velocity..
    so: v= 7t^2 -5t +3
    s= 7t^3/3 - 5t^2/2 + 3t

    ReplyDelete