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anonymous

  • one year ago

A particle moves with a constant acceleration of (0.1i-0.2j)ms^-2 it is initially at the origin where it has velocity (-i+3j)ms^-1. i and j are east and north Find the speed of the particle when it is traveling south east

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  1. Michele_Laino
    • one year ago
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    using this reference frame: |dw:1434044128319:dw| the speed of the particle is given by the subsequent equations: \[\Large \left\{ \begin{gathered} {v_x}\left( t \right) = 0.1t - 1 \hfill \\ {v_y}\left( t \right) = - 0.2t + 3 \hfill \\ \end{gathered} \right.\]

  2. Michele_Laino
    • one year ago
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    |dw:1434044305726:dw|

  3. Michele_Laino
    • one year ago
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    for example, at t=10 seconds, we have: \[\Large \left\{ \begin{gathered} {v_x}\left( {10} \right) = 0.1 \times 10 - 1 = 0 \hfill \\ {v_y}\left( {10} \right) = - 0.2 \times 10 + 3 = 1 \hfill \\ \end{gathered} \right.\]

  4. anonymous
    • one year ago
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    Thanks, this is using a simplified version of the equation v=u+at?

  5. anonymous
    • one year ago
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    and would it be correct to say that when it is travelling south east i is equal to -j ?

  6. Michele_Laino
    • one year ago
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    yes! I have used that formula, and south east means that our particle has the speed oriented as -j and i

  7. anonymous
    • one year ago
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    Thanks i got the answer, this helped a lot.

  8. anonymous
    • one year ago
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    Aww isn't this adorable

  9. anonymous
    • one year ago
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    nice wolf

  10. anonymous
    • one year ago
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    Say Robdogg, did you ever clear up them herpes?

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