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Math Help - S U V A T, please help.

  1. #1
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    S U V A T, please help.

    Hi guys,

    I have the following problem:

    At time = 0, a body is projected from an origin O with an initial velocity of 10ms-. The body moves along a straight line with a constant acceleration of -2ms-.

    a) Find the displacement of the body from O when t = 7.
    I have done this OK

    b) How far from O does the body come to a instantaneous rest and what is the value of t then?
    I have done this OK

    c) Find the distance travelled by the body during the time interval t = 0 to t = 7.
    I do not know this, help please
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  2. #2
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    Where are told,
    a(t)=-2\frac{\mbox{m}}{\mbox{sec}^2}
    The the velocity is the integral of that,
    v(t)=-2t+C
    But the initial conditions say,
    v(0)=10
    Thus,
    v(t)=-2t+10

    So the distance is,
    \int_0^7 -2t+10dt=-t^2+10t|_0^7=-49+70=21

    The displacement is,
    \int_0^7 |-2t+10| dt
    This integral is more confusing to integrate.
    But the trick is to divide it into parts.
    One where the -2t+10>0 and the other when -2t+10<0
    When does this change happen?
    Simple find,
    -2t+10<0
    -2t<-10
    t>5
    So at t=5 we have a change.
    Thus,
    \int_0^7 |-2t+10|dt=\int_0^5 |-2t+10|dt+\int_5^7 |-2t+10|dt
    But because of the changing signs we can write,
    \int_0^5 -2t+10dt+\int_5^7 10-2t dt
    Thus,
    -t^2+10t |_0^5 +10t-t^2|_5^7
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  3. #3
    Forum Admin topsquark's Avatar
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    Quote Originally Posted by ThePerfectHacker View Post
    Where are told,
    a(t)=-2\frac{\mbox{m}}{\mbox{sec}^2}
    The the velocity is the integral of that,
    v(t)=-2t+C
    But the initial conditions say,
    v(0)=10
    Thus,
    v(t)=-2t+10

    So the distance is,
    \int_0^7 -2t+10dt=-t^2+10t|_0^7=-49+70=21

    The displacement is,
    \int_0^7 |-2t+10| dt
    This integral is more confusing to integrate.
    But the trick is to divide it into parts.
    One where the -2t+10>0 and the other when -2t+10<0
    When does this change happen?
    Simple find,
    -2t+10<0
    -2t<-10
    t>5
    So at t=5 we have a change.
    Thus,
    \int_0^7 |-2t+10|dt=\int_0^5 |-2t+10|dt+\int_5^7 |-2t+10|dt
    But because of the changing signs we can write,
    \int_0^5 -2t+10dt+\int_5^7 10-2t dt
    Thus,
    -t^2+10t |_0^5 +10t-t^2|_5^7
    Everything ThePerfectHacker said was correct,

    EXCEPT

    he's got the roles of distance and displacement reversed.
    \int_0^7 -2t+10dt <--Gives the displacement.

    \int_0^7 |-2t+10| dt<--Gives the distance.

    -Dan
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