Lesson Walking Speed, Time, and Distance
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Source code of 'Walking Speed, Time, and Distance'
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<b>Problem:</b> The moving sidewalk at Chicago's O'Hare International Airport moves at a speed of 2.0 feet per second. Walking on the moving sidewalk, Nancy Killian walks 120 feet in the same time that it takes her to walk 52 feet without the moving sidewalk. How fast does Nancy walk? . <b>Solution:</b> Distance problems always use the fundamental distance equation: . d = r*t, where d=distance, r=rate, and t=time. . In this case we have two distances and one rate. . d = 120 r = w+2, where 'w' is Nancy's walking rate and 2 ft/sec is the speed of the moving sidewalk. . d<sub>1</sub> = 52 r<sub>1</sub> = w, where 'w' is Nancy's walking rate without an assist from a moving sidewalk. . At this point, we have too many unknowns to solve the equation. But we have one addition bit of information that is very helpful: it takes the same time. So, t = t<sub>1</sub> or rather t=t. That indicates we need to get both distance equations into t=something form to solve the problem. . <i>With Moving Sidewalk</i> . d = r*t d = 120 r = w+2 t = 120/(w+2) . <i>Without Moving Sidewalk</i> . d<sub>1</sub> = 52 r<sub>1</sub > = w t<sub>1</sub> = 52/w . <i>Transitive Property of Equality</i> . We know t=t<sub>1</sub>, so that means: . 120/(w+2) = 52/w . Cross-multiply. . 120w = 52(w+2) . 120w = 52w + 104 . Subtract 52w from both sides. . 120w -52w = (52w-52w) + 104 = 104 . 68w = 104 . Divide both sides by 68. . w = 104/68 = 26/17 . This solution indicates Nancy's walking speed is 26/17 ft/sec or approximately 1.529411764705882 ft/sec. . But <i>always</i> check your proposed solution to ensure it is the answer. . If w=1.529411764705882, how long does it take her to walk 52 ft? 52/1.529411764705882 = 34.00000000000001 sec. . If w+2 = 3.529411764705882, how long does it take her to walk 120 ft? 120/3.529411764705882 = 34.00000000000001 sec. . Correct. . <b>Answer:</b> Nancy's walking speed is 26/17 ft/sec or approximately 1.529411764705882 ft/sec.
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