#### New-Blu-Blood

1. The problem statement, all variables and given/known data

Initially, a particle is moving at 4.40 m/s at an angle of 35.0 degrees above the horizontal. Two seconds later, its velocity is 6.35 m/s at an angle of 58.0 degrees below the horizontal. What was the particle's average acceleration during these 2.00 seconds in terms of the components of the average acceleration?

2. Relevant equations[/

[]3. The attempt at a solution

Ok, I found the x component of the first given vector to be 4.40 x cos(35)=3.604. Then i found the y component 4.40 x sin(35)=2.5237. I then found the components of the second given vector to be 6.35 x cos(-58)= 3.3649 for the x component and for the y component I found 6.35 x sin(-58) = -5.285.

I added 3.604 and 3.3649 and divided that by the time=2 seconds to get the acceleration x vector of 6.969. Then i added 2.5237 and -5.385 and divided that sum by 2 to get -2.86 as my acceleration y vector.

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#### rl.bhat

Homework Helper
Hi New-Blu-Blood, welcome to PF.
Show your calculation.Then we will find out where you went wrong?

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