Average acceleration of particle

[wick85]

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?

 

[rl.bhat]

To evaluate the change in the velocities, reverse the direction of the initial velocity and then find the components along x and y axis.

 

[tomcue]

As a scientist, it is important to carefully analyze the given information and use appropriate equations and calculations to determine the correct answer. In this case, the given information includes initial and final velocities at different angles, as well as a specific time interval. To find the average acceleration, we can use the formula Aav = (Vf - Vi)/T, where Vf is the final velocity, Vi is the initial velocity, and T is the time interval.

In the x-direction, we can use the initial and final velocities at the given angles to calculate the components of the velocities in the x-direction using the equations Vix = Vo cos theta. Using the given information, we can calculate Vix for both the initial and final velocities. Plugging these values into the average acceleration formula, we get:

Aav,x = (Vfx - Vix)/T = (3.38 m/s - 4.27 m/s)/2 s = -0.445 m/s^2

In the y-direction, we can use the same process to find the average acceleration. Using the equation Viy = Vo sin theta, we can calculate the initial and final velocities in the y-direction. Plugging these values into the average acceleration formula, we get:

Aav,y = (Vfy - Viy)/T = (5.28 m/s - 3.09 m/s)/2 s = 1.095 m/s^2

Therefore, the particle's average acceleration during the 2.00 seconds is -0.445 m/s^2 in the x-direction and 1.095 m/s^2 in the y-direction. These values represent the change in velocity per unit time in each direction, and they can be used to further analyze the motion of the particle. It is important to carefully consider the given information and use the appropriate equations and calculations to accurately determine the average acceleration.