anonymous
  • anonymous
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
Physics
  • Stacey Warren - Expert brainly.com
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SOLVED
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chestercat
  • chestercat
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Michele_Laino
  • Michele_Laino
using this reference frame: |dw:1434044128319:dw| the speed of the particle is given by the subsequent equations: \[\Large \left\{ \begin{gathered} {v_x}\left( t \right) = 0.1t - 1 \hfill \\ {v_y}\left( t \right) = - 0.2t + 3 \hfill \\ \end{gathered} \right.\]
Michele_Laino
  • Michele_Laino
|dw:1434044305726:dw|
Michele_Laino
  • Michele_Laino
for example, at t=10 seconds, we have: \[\Large \left\{ \begin{gathered} {v_x}\left( {10} \right) = 0.1 \times 10 - 1 = 0 \hfill \\ {v_y}\left( {10} \right) = - 0.2 \times 10 + 3 = 1 \hfill \\ \end{gathered} \right.\]

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anonymous
  • anonymous
Thanks, this is using a simplified version of the equation v=u+at?
anonymous
  • anonymous
and would it be correct to say that when it is travelling south east i is equal to -j ?
Michele_Laino
  • Michele_Laino
yes! I have used that formula, and south east means that our particle has the speed oriented as -j and i
anonymous
  • anonymous
Thanks i got the answer, this helped a lot.
anonymous
  • anonymous
Aww isn't this adorable
anonymous
  • anonymous
nice wolf
anonymous
  • anonymous
Say Robdogg, did you ever clear up them herpes?

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