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 40.3 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 18.1 s to reach the end of the walkway. Then, she turns around and walks as before with the same speed relative to the conveyor, just this time in the opposite direction. The return leg takes 68.3 s. a) What is the speed of the walkway conveyor relative to the terminal? b) With what speed was the girl walking?

Explanation / Answer

Let the speed with which the girl walks be v and that with which the conveyor moves be u. In the first case the total speed for the girl will be v + u , as she is moving along the direction of motion of conveyor. so v+ u= d/t = 42.5/14.1 = 3.01m/s in the second case the net speed of the girl is v-u = d/t = 42.5/72.7 = 0.58 m/s adding the two 2v = 3.59 => v = 2.79 m/s . That is the speed of girl. using it in any of the two equations, we will get the speed of the conveyor, which will come out to be 0.22 m/s

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