During a long airport layover, a physicist father and his 8-year-old daughter tr

Question

During a long airport layover, a physicist father and his 8-year-old daughter try a game that involves a moving walkway. They have measured the walkway to be 47.6 m long. The father has a stopwatch and times his daughter. First, the daughter walks with a constant speed in the same direction as the conveyor. It takes 15.0 s to reach the end of the walkway. Then, she turns around and walks with the same speed relative to the conveyor as before in the opposite direction. The return leg takes 71.4 s. What is the speed of the walkway conveyor relative to the terminal, and with what speed was the girl walking?

Explanation / Answer

To solve this problem, you can setup the following equations: (Vw + Vg) * 15 seconds = 47.6 m where Vw is walkway speed and Vg is the girl's speed (Vw - Vg) * 71.4 seconds = 47.6m If you solve for Vw in the first equation, you get: 15Vw = 47.6 - 15Vg Vw = 3.17 - Vg Now, if you plug that into the second equation in place of the Vw, you can solve for Vg: ((3.17 - Vg) - Vg) * 71.4 = 47.6 226.34 - 142.8Vg = 47.6 178.74 = 142.8Vg Vg = 1.25 m/s Now that you know Vg, you can solve for Vw easily: Vw = 3.17 - Vg Vw = 3.17 - 1.25 Vw = 1.92 m/s

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