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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

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

|dw:1434044305726:dw|

3. Michele_Laino

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

4. anonymous

Thanks, this is using a simplified version of the equation v=u+at?

5. anonymous

and would it be correct to say that when it is travelling south east i is equal to -j ?

6. Michele_Laino

yes! I have used that formula, and south east means that our particle has the speed oriented as -j and i

7. anonymous

Thanks i got the answer, this helped a lot.

8. anonymous

Aww isn't this adorable

9. anonymous

nice wolf

10. anonymous

Say Robdogg, did you ever clear up them herpes?

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