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

Question

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

michele_laino · Answer

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 · Answer

|dw:1434044305726:dw|

michele_laino · Answer

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

anonymous · Answer

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

anonymous · Answer

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

michele_laino · Answer

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

anonymous · Answer

Thanks i got the answer, this helped a lot.

anonymous · Answer

Aww isn't this adorable

anonymous · Answer

nice wolf

anonymous · Answer

Say Robdogg, did you ever clear up them herpes?