Initially, a particle is moving at 4.00 m/s at an angle of 33.5 degrees above the horizontal. Two seconds later, its velocity is 6.45 m/s at an angle of 60.0 degrees below the horizontal. What was the particle's average acceleration during these 2.00 seconds? Enter the x and y components of the average acceleration separated by comma.
Aavgx= ?
Aavgy= ?
I don't have a clue where to start.
never mnind, i just saw similar posts to my question
To find the average acceleration, we need to find the change in velocity and divide it by the time interval. The change in velocity can be found by subtracting the initial velocity from the final velocity.
Given:
Initial velocity (v1) = 4.00 m/s at an angle of 33.5 degrees above the horizontal
Final velocity (v2) = 6.45 m/s at an angle of 60.0 degrees below the horizontal
Time interval (t) = 2.00 seconds
1. Convert the velocities to their horizontal (x) and vertical (y) components.
To do this, we will use trigonometry.
For the initial velocity (v1):
vx1 = v1 * cos(angle1)
vy1 = v1 * sin(angle1)
For the final velocity (v2):
vx2 = v2 * cos(angle2)
vy2 = v2 * sin(angle2)
Substituting the given values:
vx1 = 4.00 * cos(33.5°)
vy1 = 4.00 * sin(33.5°)
vx2 = 6.45 * cos(-60.0°)
vy2 = 6.45 * sin(-60.0°)
2. Calculate the change in velocity for both the x and y components.
Δvx = vx2 - vx1
Δvy = vy2 - vy1
3. Calculate the average acceleration for each component.
Aavgx = Δvx / t
Aavgy = Δvy / t
Now, let's calculate the values step-by-step:
Step 1:
vx1 = 4.00 * cos(33.5°) ≈ 3.365 m/s
vy1 = 4.00 * sin(33.5°) ≈ 2.143 m/s
vx2 = 6.45 * cos(-60.0°) ≈ -3.225 m/s
vy2 = 6.45 * sin(-60.0°) ≈ -5.577 m/s
Step 2:
Δvx = vx2 - vx1 ≈ -3.225 - 3.365 ≈ -6.590 m/s
Δvy = vy2 - vy1 ≈ -5.577 - 2.143 ≈ -7.720 m/s
Step 3:
Aavgx = Δvx / t ≈ -6.590 / 2.00 ≈ -3.295 m/s^2
Aavgy = Δvy / t ≈ -7.720 / 2.00 ≈ -3.860 m/s^2
Therefore, the x-component of the average acceleration (Aavgx) is approximately -3.295 m/s^2, and the y-component of the average acceleration (Aavgy) is approximately -3.860 m/s^2.