Required information A woman of mass $52.3 \mathrm{~kg}$ is standing in an elevator. If the elevator maintains constant acceleration and is moving at $1.50 \mathrm{~m} / \mathrm{s}$ as it passes the fourth floor on its way down, what is its speed 4.00 s later? $\mathrm{m} / \mathrm{s}$

Step 1 ：We are given that a woman of mass 52.3 kg is standing in an elevator. The elevator maintains constant acceleration and is moving at 1.50 m/s as it passes the fourth floor on its way down. We are asked to find its speed 4.00 s later.

Step 2 ：We know that the elevator is moving at a constant acceleration. We also know the initial velocity of the elevator and the time after which we need to find the final velocity. We can use the equation of motion to find the final velocity.

Step 3 ：The equation of motion is given by: \(v = u + at\) where: \(v\) is the final velocity, \(u\) is the initial velocity, \(a\) is the acceleration, and \(t\) is the time.

Step 4 ：We know the initial velocity (\(u = 1.5\) m/s) and the time (\(t = 4.0\) s), but we don't know the acceleration (\(a\)). However, since the elevator is moving downwards, the acceleration is due to gravity, which is approximately 9.8 m/s².

Step 5 ：So, we can substitute these values into the equation to find the final velocity: \(v = u + at = 1.5 + 9.8 \times 4.0\)

Step 6 ：Calculating the above expression, we get \(v = 40.7\) m/s

Step 7 ：Final Answer: The speed of the elevator 4.00 s later is \(\boxed{40.7 \, \mathrm{m/s}}\)