An escalator that has 30 steps moves upward at the rate of one step per second. Franklin gets on and walks up at his own speed. At the top, he realizes that he dropped a penny on the lower floor and believing that "a penny saved is a penny earned" decides to walk back down and retrieve it. He uses the same escalator so that he can keep the penny in view. Franklin walks down at the same speed that he walked up. It takes him twice as long to get down as it took him to go up. How fast is his walking speed?
The 'distance' to travel is 30 steps. The speed up is (his speed=u) u+1, the speed down, u-1. So, t=distance/speed \[2\frac{30}{u+1}=\frac{30}{u-1}\],as the time down (right side) is twice the speed up. \[\frac{60u-60}{u+1}=30\] \[60u-60=30u+30\] \[30u-60=30\] \[30u=90\] \[u=3 (steps/second) \]
Which makes sense, as the speed up (3+1) is twice the speed down (3-1)
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