An elevator is initially moving upward at a speed of 12.84 m/s. The elevator experiences a constant downward acceleration of magnitude 3.80 m/s2 for 3.18 s.

(a) Find the magnitude and direction of the elevator's final velocity.
magnitude:
direction:

To find the magnitude and direction of the elevator's final velocity, we can use the kinematic equation:

vf = vi + at

where vf is the final velocity, vi is the initial velocity, a is the acceleration, and t is the time.

Given:
Initial velocity (vi) = 12.84 m/s (upward)
Acceleration (a) = 3.80 m/s^2 (downward)
Time (t) = 3.18 s

First, let's find the change in velocity due to acceleration:

Δv = a * t

Δv = 3.80 m/s^2 * 3.18 s
Δv = 12.084 m/s (downward)

Now, let's find the final velocity:
To do this, we need to consider both the initial velocity and the change in velocity due to acceleration.

If the acceleration is in the opposite direction as the initial velocity, their magnitudes subtract. If they are in the same direction, their magnitudes add.

Since the initial velocity is upward and the acceleration is downward, their magnitudes subtract.

vf = vi - Δv

vf = 12.84 m/s (upward) - 12.084 m/s (downward)
vf = 0.756 m/s (upward)

So, the magnitude of the elevator's final velocity is 0.756 m/s and its direction is upward.