Initially, a particle is moving at 5.27 m/s at an angle of 35.9° above the horizontal. Two seconds later, its velocity is 6.27 m/s at an angle of 57.4° below the horizontal. What was the particle's average acceleration during these 2.00 seconds in the x-direction and the y-direction? 2. Relevant equations Aav = Vf -Vi/Tf-Ti Vix = Vo cos theta Viy = Vo sin theta 3. The attempt at a I drew triangles for both: for the first triangle i got Vix = 4.27 m/s and Viy 3.09 m/s For the second triangle I got Vix = 3.38 m/s and Viy 5.28 m/s I got an answer of -0.445 m/s/s (x direction) and 1.09 m/s/s (y direction) but apparently those arent the right answers...Any suggestions?