24. An airplane, travelling initially at 240 m/s [28° S of
E], takes 35 s to change its velocity to 220 m/s [28° E
of S]. What is the average acceleration over this time
interval?
To find the average acceleration, we first need to find the change in velocity and the time interval.
The change in velocity can be calculated by finding the vector difference between the two velocities. We can break down the initial and final velocities into their x and y components:
Initial velocity (240 m/s, 28° S of E):
Vx = 240 * cos(28°) = 213.312 m/s
Vy = 240 * sin(28°) = -111.103 m/s
Final velocity (220 m/s, 28° E of S):
Vx = 220 * sin(28°) = 97.237 m/s
Vy = 220 * cos(28°) = 194.156 m/s
Now we can calculate the change in velocity:
ΔVx = 97.237 - 213.312 = -116.075 m/s
ΔVy = 194.156 - (-111.103) = 305.259 m/s
ΔV = sqrt(ΔVx^2 + ΔVy^2) = sqrt((-116.075)^2 + (305.259)^2) = 329.696 m/s
The time interval is given as 35 s.
Now, we can calculate the average acceleration:
Average acceleration = ΔV / Δt
Average acceleration = 329.696 / 35
Average acceleration ≈ 9.42 m/s^2
Therefore, the average acceleration over this time interval is approximately 9.42 m/s^2.