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Math Help - A level Maths 05- cont.

  1. #1
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    Thumbs up A level Maths 05- cont.

    this is the rest of the homework .thank you
    Attached Thumbnails Attached Thumbnails A level Maths 05- cont.-untitled-5.jpg   A level Maths 05- cont.-untitled-6.jpg  
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  2. #2
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    Hello, carlasader!

    Here's #7 . . .


    7. Two cars A and B are moving on straight horizontal roads with constant velocities.
    The velocity of A is 20 m/sec due east, the velocity of B is (10i + 10j) m/sec,
    where i and j are unit vectors directed due east and due north, respectively.
    Initially A is at the origin O, and the position vector of B is 300i.
    At time t seconds, the position vectors of A and B are \vec{r} and \vec{s}, resp.

    a) Find expressions for \vec{r} and \vec{s} in terms of t.
    \vec{r}\:=\:(20t)i
    \vec{s}\:=\:(300 + 10t)i + (10t)j


    b) Hence write an expression for \overrightarrow{AB} in terms is t.
    We have: . \vec{A} \:=\:\langle 20t,\,0\rangle,\;\vec{B}\:=\:\langle300 + 10t,\,10t\rangle

    Then: . \overrightarrow{AB} \;=\;\langle 300-10t,\:10t\rangle



    c) Find the time when the bearing of B from A is 045^o.
    Since B is always on a bearing of 45, this occurs when: A \,= \,\langle 300t,\:0\rangle
    . . Then: . \langle20t,\:0\rangle \:=\:\langle 300,\:0\rangle\quad\Rightarrow\quad t = 15 seconds.



    d) Find the time when the cars are again 300 m apart.
    The distance is: . |AB| \;=\;\sqrt{(300 - 10t)^2 + (10t-0)^2} \;=\;\sqrt{90,000 - 6000t + 200t^2}

    If |AB| = 300:\;\;\sqrt{90,000 - 6000t + 200t^2} \:=\:300

    Square both sides: . 90,000 - 6000t + 200t^2\:=\:90,000\quad\Rightarrow\quad 200t^2 - 6000t \:=\:0

    Factor: . 200t(t - 30) \:=\:0\quad\Rightarrow\quad t \,=\,0,\:30


    Therefore, they are 300 m apart (again) when t = 30 seconds.

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  3. #3
    Grand Panjandrum
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    a) As the forces acting on B are -3mg and T, and the accelleration of the
    particle is -(1/2)g, the tension must be (1/2)3mg. That is we have:

    -3mg + T = 3ma,

    so:

    T = 3ma + 3mg

    but a= -(1/2)g.

    b) Resolve the gravitational force on A into components up the plene and
    normal to the plane.

    The component normal to the plane is -mg sin(theta), and up the plane is
    -mg cos(theta), also sin(theat)=3/5, and cos(theta)=4/5.

    Now the forces on the particle up the plane are T = (3/2)mg, the component
    of the weight acting up the plane -(4/5)mg, and the friction -mu (3/5)mg

    So the total force up the plane is:

    T - (4/5)mg - mu (3/5)mg

    and the acceleration up the plane is (1/2)g, so:

    (3/2)mg - (4/5)mg - mu (3/5)mg = (1/2) mg

    or cancelling mg:

    (3/2) - (4/5) - mu (3/5) = (1/2)

    Which I will leave to the reader to finish.

    RonL
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