Vector Problem: Avg Acc Components

[New-Blu-Blood]

Homework Statement

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. Homework 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.

 

[rl.bhat]

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

 

[tomcue]

To find the magnitude of the average acceleration, we can use the Pythagorean theorem:

Average acceleration = √(6.969^2 + (-2.86)^2) = 7.73 m/s^2

The direction of the average acceleration can be found using trigonometric functions:

tanθ = (-2.86)/(6.969)

θ = tan^-1(-2.86/6.969) = -22.5 degrees

Therefore, the average acceleration during these 2.00 seconds is 7.73 m/s^2 at an angle of -22.5 degrees below the horizontal.