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Average acceleration of particle

  • Thread starter wick85
  • Start date
  • #1
4
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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?



Homework 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?
 

Answers and Replies

  • #2
rl.bhat
Homework Helper
4,433
7
To evaluate the change in the velocities, reverse the direction of the initial velocity and then find the components along x and y axis.
 

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