A spotlight on the ground is shining on a wall $24 \mathrm{~m}$ away. If a woman $2 \mathrm{~m}$ tall walks from the spotlight toward the building at a speed of $0.6 \mathrm{~m} / \mathrm{s}$, how fast is the length of her shadow on the building decreasing when she is $4 \mathrm{~m}$ from the building?
Answer (in meters per second): 1.8

Step 1 ：Let x be the distance between the woman and the building, and let y be the length of the shadow on the building. We have the following similar triangles: \(\frac{2}{x} = \frac{y}{x + 24}\)

Step 2 ：Find the length of the shadow when the woman is 4 meters away from the building: \(\frac{2}{4} = \frac{y}{4 + 24}\) which gives y = 14

Step 3 ：Differentiate both sides with respect to time t: \(-\frac{2}{x^2} \frac{dx}{dt} = \frac{\frac{dy}{dt}(x + 24) - y \frac{dx}{dt}}{(x + 24)^2}\)

Step 4 ：Plug in the values we know (x = 4, y = 14, dx/dt = 0.6) and solve for dy/dt: \(-\frac{2}{4^2} \cdot 0.6 = \frac{\frac{dy}{dt}(4 + 24) - 14 \cdot 0.6}{(4 + 24)^2}\)

Step 5 ：Solve for dy/dt: dy/dt = -1.8

Step 6 ：\(\boxed{\text{Final Answer: The length of the woman's shadow on the building is decreasing at a rate of 1.8 meters per second when she is 4 meters away from the building.}}\)