MATH SOLVE

6 months ago

Q:
# A rope, attached to a weight, goes up through a pulley at the ceiling and back down to a worker. The worker holds the rope at the same height as the connection point between the rope and weight. The distance from the connection point to the ceiling is 40 ft. Suppose the worker stands directly next to the weight (i.e., a total rope length of 80 ft) and begins to walk away at a constant rate of 2 ft/s. How fast is the weight rising when the worker has walked: 10 feet? Answer = 30 feet? Answer =

Accepted Solution

A:

Answer: 0.66 ftStep-by-step explanation:Let assume that the initial position of the worker is x. Given that the worker walks away with a constant speed of 2 ft/s. Therefore, dx/dt = 2As the worker moves away, the rope makes a triangle, with width length x and the height length will be 30.Using pythagorean theorem, the length of rope on this side of the pulley will be √(x² + 30²)Also, the length of rope on the other side will be 60 - √(x² + 30²), and the height h of the weight will be 30 - (60 - √(x² + 30²)) = √(x² + 30²) - 30dh/dt = dx/dt × x/√(x² + 30²) = 4x/√(x² + 30²)dh/dt = 4x/√(x² + 30²)If the worker moves 5ft away, then dh/dt = (4×5)/√(5² + 30²) dh/dt = 20/√(25 + 900) dh/dt = 0.66 ft