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# solution_2pdf - armington(kma786 hw0127 ete(57165 This...

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armington (kma786) – hw0127 – fete – (57165) 1 This print-out should have 10 questions. Multiple-choice questions may continue on the next column or page – fnd all choices beFore answering. 001 10.0 points A car travels along a straight stretch oF road. It proceeds For 16 . 2 mi at 56 mi / h, then 20 . 8 mi at 44 mi / h, and fnally 49 . 8 mi at 34 . 7 mi / h. What is the car’s average velocity during the entire trip? Correct answer: 39 . 5053 mi / h. Explanation: Let : d A = 16 . 2 mi , v A = 56 mi / h , d B = 20 . 8 mi , v B = 44 mi / h , d C = 49 . 8 mi , and v C = 34 . 7 mi / h . The total time the car spent on the road is Δ t = d A v A + d B v B + d C v C = 16 . 2 mi 56 mi / h + 20 . 8 mi 44 mi / h + 49 . 8 mi 34 . 7 mi / h = 2 . 19717 h , so the average velocity is v = Δ d Δ t = d A + d B + d C Δ t = 16 . 2 mi + 20 . 8 mi + 49 . 8 mi 2 . 19717 h = 39 . 5053 mi / h . 002 10.0 points A car makes a 244 km trip at an average speed oF 40 . 2 km / h. A second car starting 1 h later arrives at their mutual destination at the same time. What was the average speed oF the second car For the period that it was in motion? Correct answer: 48 . 1295 km / h. Explanation: Let : d = 244 km , v = 40 . 2 km / h , and Δ t = 1 h . Average velocity is v = d t . Δ t = t 1 - t 2 = d v 1 - d v 2 d v 2 = d v 1 - Δ t v 2 d = 1 d v 1 - Δ t · v 1 v 1 v 2 = d v 1 d - v 1 Δ t = (244 km)(40 . 2 km / h) 244 km - (40 . 2 km / h)(1 h) = 48 . 1295 km / h . 003 (part 1 of 2) 10.0 points A runner is jogging at a steady 3 . 1 km / hr. When the runner is 4 . 4 km From the fnish line, a bird begins ±ying From the runner to the fnish line at 9 . 3 km / hr (3 times as Fast as the runner). When the bird reaches the fnish line, it turns around and ±ies back to the runner.

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solution_2pdf - armington(kma786 hw0127 ete(57165 This...

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