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Math Help - position vector 2

  1. #1
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    position vector 2

    At any time t , a particle has a position vector (t^2+1)i+(3t-2)j , relative to the origin O . Show that the acceleration is always directed horizontally . Hence , find the angle between the velocity and acceleration of the particle at any time , t .

    v=2ti+3j , a=2i

    Since a=2i , the particle is always accelerating along the x-axis .

    \cos \theta=\frac{2t(2)+3(0)}{\sqrt{4t^2+9}\cdot \sqrt{4}}

    =\frac{2t}{\sqrt{4t^2+9}}

    so am i correct ?
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  2. #2
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    Quote Originally Posted by hooke View Post
    At any time t , a particle has a position vector (t^2+1)i+(3t-2)j , relative to the origin O . Show that the acceleration is always directed horizontally . Hence , find the angle between the velocity and acceleration of the particle at any time , t .

    v=2ti+3j , a=2i

    Since a=2i , the particle is always accelerating along the x-axis .

    \cos \theta=\frac{2t(2)+3(0)}{\sqrt{4t^2+9}\cdot \sqrt{4}}

    =\frac{2t}{\sqrt{4t^2+9}}

    so am i correct ?
    seems fine
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