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M2 Mechanics EdExcel Review Exercise 1 Question watch


    Hi, I am really stuck on this M2 question and I do not get how its possible to work out the answer. If you have the M2 book,its Review Exercise 1 page 85 question 10b.

    Here is the question 10b) if u dont have the M2 book,

    Key:-
    =< means greater than or equal to
    <= means less than or equal to

    10b) A particle P moves in a straight line in such a way that, at time t seconds, its velocity v m/s is given by:

    v=12t-3t^2 , for 0=<t<=5

    v= (-375)t^(-2), for t>5

    When t=0, P is at the point 0. Calculate the displacement of P from 0 when t=6.

    [To start this question integrate v gives displacement x]
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    2
    ReputationRep:
    Int[0,X]1dx = Int[0,5] 12t - 3t^2 dt
    => X = [6t^2 - t^3] between[0,5] = 25

    therefore Int[25,Y] dx = Int[5,6] -375t^-2 dt
    => Y - 25 = 375/6 - 375/5 => Y = 25/2
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    2
    ReputationRep:
    v=ds/dt
    ds/dt=12t-3t^2
    integrating,
    s=6t^2 - t^3 + C => C=0 @ t=0
    i.e. s=6t^2 - t^3, 0<=t<=5

    @t=5,
    s=6(25) - (125)=150-125=25
    s=25m

    also,
    ds/dt=-375t^(-2), t>5
    integrating,
    (oops)
    s=375t^(-1) + D

    @t=5,
    25=375/(5) + D
    25=75+D
    D=-50
    s=375/t - 50, t>5

    @t=6,
    s=375/(6)-50=62.5-50
    s=12.5m
 
 
 
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