Initially, a particle is moving at 4.20m/s at an angle of 34.5∘ above the horizontal. Two seconds later, its velocity is 6.50m/s at an angle of 50.0∘ below the horizontal.
What was the particle's average acceleration during these 2.00 seconds?
change in x speed = 6.5 cos50-4.2cos34.5
change in y speed = -6.5 sin50-4.2sin34.5
ax = change in x speed/2
ay = change in y speed/2
a = sqrt (ax^2+ay^2)
A = angle of a above horizontal
tan A = ay/ax
To find the particle's average acceleration during the 2.00 seconds, we need to determine the change in velocity and divide it by the change in time.
1. First, we need to find the initial velocity and final velocity components in the horizontal and vertical directions.
Initial horizontal velocity:
Vix = Vinitial * cos(theta_initial)
= 4.20 m/s * cos(34.5°)
Initial vertical velocity:
Viy = Vinitial * sin(theta_initial)
= 4.20 m/s * sin(34.5°)
Final horizontal velocity:
Vfx = Vfinal * cos(theta_final)
= 6.50 m/s * cos(-50.0°) (since the angle is below the horizontal)
Final vertical velocity:
Vfy = Vfinal * sin(theta_final)
= 6.50 m/s * sin(-50.0°)
Note: The negative sign on the angle is because the angle is below the horizontal.
2. Next, calculate the change in velocity for horizontal and vertical components:
Change in horizontal velocity:
ΔVx = Vfx - Vix
Change in vertical velocity:
ΔVy = Vfy - Viy
3. Lastly, divide the change in velocity by the change in time:
Average acceleration = ΔV / Δt
Average acceleration in the horizontal direction:
Ax = ΔVx / Δt
Average acceleration in the vertical direction:
Ay = ΔVy / Δt
The average acceleration during the 2.00 seconds is given by the magnitude of the average acceleration vector:
Average acceleration = √(Ax^2 + Ay^2)