An elevator is initially moving upward at a speed of 11.04 m/s. The elevator experiences a constant downward acceleration of magnitude 3.84 m/s2 for 3.06 s.

(a) Find the magnitude and direction of the elevator's final velocity.
(b)How far did it move during the 3.06 s interval?

To find the magnitude and direction of the elevator's final velocity, we can use the following equations of motion:

1. Final velocity (v) = Initial velocity (u) + (acceleration * time)
2. Final velocity squared (v^2) = Initial velocity squared (u^2) + 2 * acceleration * distance

(a) To find the magnitude and direction of the elevator's final velocity, we can use equation (1) with the given values:

Initial velocity (u) = 11.04 m/s (upward, chosen as positive)
Acceleration (a) = -3.84 m/s^2 (downward, chosen as negative)
Time (t) = 3.06 s

Plugging in these values into equation (1), we get:
Final velocity = 11.04 m/s + (-3.84 m/s^2 * 3.06 s)
Final velocity = 11.04 m/s + (-11.7504 m/s)
Final velocity = -0.7104 m/s

Therefore, the magnitude of the elevator's final velocity is 0.7104 m/s, and since it is negative, the direction is downward.

(b) To find the distance the elevator moved during the 3.06 s interval, we can use equation (2) with the given values:

Initial velocity (u) = 11.04 m/s
Acceleration (a) = -3.84 m/s^2
Time (t) = 3.06 s

Since we want to find the distance, we rearrange equation (2) to solve for distance:

Distance (d) = (Final velocity squared - Initial velocity squared) / (2 * acceleration)

Plugging in the values into the equation, we get:
Distance = ((-0.7104 m/s)^2 - (11.04 m/s)^2) / (2 * (-3.84 m/s^2))
Distance = (0.504144 m^2/s^2 - 121.3824 m^2/s^2) / (-7.68 m/s^2)
Distance = (-120.878256 m^2/s^2) / (-7.68 m/s^2)
Distance = 15.7285 m

Therefore, the elevator moved 15.7285 meters during the 3.06 second interval.