MAT-1020 Final: Measurements and Units

25. Two carts, A and B, are connected by a rope 39 ft long that passes over a pulley P (see the figure). The point Q is on the floor 12 ft directly beneath P and between the carts. Cart A is being pulled away from at a speed of 2 ft/s. How fast is cart B moving toward at the instant when cart A is 5 ft from 0? 12ft 0

A swimming pool is 20 ft wide, 40 ft long, 3 ft deep at the shallow end, and 9 ft deep at is deepest point. A cross-section is shown in the figure. If the pool is being filled at a rate of 1.2 ft3/min, how fast is the water level rising when the depth at the deepest point is 5 ft? (Round your answer to five decimal places.) ft/min Gravel is being dumped from a conveyor belt at a rate of 40 ft3/min, and its coarseness is such that it forms a pile in the shape of a cone whose base diameter and height are always equal. How fast is the height of the pile increasing when the pile is 5 ft high? (Round your answer to two decimal places.) ft/min A kite 100 ft above the ground moves horizontally at a speed of 8 ft/s. At what rate is the angle (in radians) between the string and the horizontal decreasing when 200 ft of string have been let out? rad/s

EXAMPLE 4 Suppose the odometer on our car is broken and we want to estimate the distance driven over a 30 second time interval. We take the speedometer readings every five seconds and record them in the following table Time (s) 0 5 10 15 20 25 30 Velocity (mi/h) 17 21 |24 27 32 31 27 In order to have the time and the velocity in consistent units, let's convert the velocity readings to feet per second 5280 3600 (1 mi/h = ft/s). (Round your answers to the nearest whole number.) Time (s) 0 10 15 20 25 30 Velocity (ft/s) 25 35 47 45 40 During the first five seconds the velocity doesn't change very much, so we can estimate the distance traveled during that time by assuming that the velocity is constant. If we take the velocity during that time interval to be the initial velocity (25 ft/s), then we obtain the approximate distance traveled during the first five seconds 25 ft/s × 5 s- ft Similarly, during the second time interval the velocity is approximately constant and we take it to be the velocity when t = 5 s. So our estimate for the distance traveled from t = 5 s to t = 10 s is 31ft/s × 5 s- ft If we add similar estimates for the other time intervals, we obtain an estimate for the total distance traveled (25 × 5) + (31 × 5) + (35 × 5) + (40 × 5) + (47 × 5) + (45 × 5) = ft. We could just as wel have used the velocity at the end of each time period instead of the velocity at the beginning as our assumed constant velocity. Then our estimate becomes (31 × 5) + (35 × 5) + (40 × 5) + (47 × 5) + (45 × 5) + (40 × 5) = ft. If we had wanted a more accurate estimate, we could have taken velocity readings every two seconds, or even every second