Sunday, April 12, 2009

Velocity

A particle moves in a straight line so that at time t seconds after passing through a fixed point, its velocity, v m/s, is given by v = 6 cos 2t. Find
  1. the two smallest positive values of t for which the particle is at instanteneous rest
  2. the distance between the positions of instantaneous rest corresponding to these two values of t
  3. the greatest magnitude of the accelerstion
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6 comments:

  1. poopsiepumkinpieApril 13, 2009 at 11:55 AM

    omg.....where to start...ok i lie...how to start....lol sigh

    ReplyDelete
  2. poopsiepumkinpieApril 13, 2009 at 12:17 PM

    v= 6 cos 2t

    at inst rest v=0

    6 cos 2t= 0

    cos 2t= 0

    2t= cos^-1 0

    2t= 90

    t =45

    ReplyDelete
  3. poopsiepumkinpieApril 13, 2009 at 12:18 PM

    i dont no if this is correct....if not correct it plz....

    ReplyDelete
  4. Swanky ThingApril 15, 2009 at 11:53 AM

    Well am not sure how to do this question but i think you have to find dv/dt first and them find the miniumum value of it. v=o fo find for t

    ReplyDelete
  5. BladeApril 17, 2009 at 11:04 PM

    well this is a velocity qu so just as swanky thing said you have to differentiate by finding dv/dt and then find when v=o for a min value though not sure how to find the second smallest value..............mabye it's when v=0.001.........ahh help!

    ReplyDelete
  6. animechicApril 18, 2009 at 8:43 AM

    can some1 plz explain this question 4 me. i'm lost

    ReplyDelete

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