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At t = 0, a particle leaves the origin with a velocity of 9 m/s in the positive y direction and moves in the xy plane with a constant acceleration of (-2 i -4 j) m/s2. At t = 1 s, what is the velocity of the particle in the x direction?

I need explanation for this question !

  • physics -

    There is not much explaination to do, it is a vector equation.

    using i,j coordinates

    v(t)=vi+at=0i+9j +(-2t i)+ (-4t j)

    put in t=1, and solve. velocity in the i direction will be let me see
    v(1)i=-2 m/s

    check that.

  • physics -

    0i come from where ?

  • physics -

    Also,9-2-4=3 !!

  • physics -

    i,j are unit vectors, in different directions. Oi means zero in the i direction, as in the initial velocity given.

    You cannot 9-2-4 as those were in different directions.

    Surely this is in your textbook.

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