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    HEY GUYS, IM STUCK ON A PAST PAPER QUESTION, BECAUSE MY TEACHERS MARK SCHEME HE GAVE ME IS DEFINATELY WRONG AND ITS CONFUSING ME. ANY HELP WOULD BE GREATLY APPRECIATED, LOTS OF GOOD REP!! THANKS:

    6) THE DISPLACEMENT OF A PARTICLE AT TIME t SECONDS AFTER IT PASSES THROUGH A FIXED POINT IS s Metres, WHERE S = 4.8t + 0.06t^2 - 0.004t^3

    iii) find the distance travelled by the particle from the point where it reaches its maximum velocity to the point where its velocity is half its' initial velocity.

    i have already found that

    v=4.8+0.12t-0.012t^2

    a=0.12-0.024t

    the value of s when a=0 is 25m (occurs at t=5)

    Please Please try and help,many thanks !
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    Which paper is it off?

    Oh hang on got it it's Jan O4

    When a=0

    0.12 -0.024t=0

    t= 5s

    Just stick 5 back into the first eqn to get...

    s=25m

    (or that's what my mark scheme says anyway ) Am I answering the question? :confused:
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    Where did you get the mark scheme from?
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    My teacher made it! (Ok so it's not really a mark scheme!)
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    Lol, Your Answering The Question Before It Ive Already Done That 1, Please Help Me Guys !!!
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    initial velocity is when t=0, v=4.8
    4.8+0.12t-0.012t²=2.4
    0.012t²-0.12t-2.4=0
    (0.12t+1.2)(0.1t-2)=0
    t≠-10, t=20

    When t=20, s=88m

    so dist travelled = 88-25=63m

 
 
 
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