phhww3 - Physics 590 Homework, Week 3 Week 3, Homework 1...

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Physics 590 Homework, Week 3 Week 3, Homework 1 Prob. 3.1.1 A bicyclist climbs a hill at 10 km/h and then descends at constant speed returning to her starting point. Her average speed for the trip is 16 km/h. What was her downhill speed? Reasoning: If the average speed for the entire trip is greater than the constant speed for the first half of the trip, then the constant speed for the second half of the trip must be greater than the average speed. v1=10km/hr * 1000m/km * hr/3600s= 2.8 m/s v average= 16km/hr * 1000m/km * hr/3600s= 4.4 m/s v average= (v1 + v2)/2 4.4 m/s = (2.8 m/s + v2)/2 8.8 m/s = 2.8m/s – v2 6.0 m/s = v2 This result is consistent with my reasoning. Downhill speed = 6.0 m/s
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Prob. 3.1.2 A train accelerates at a constant 0.68 m/s 2 , starting from an initial speed of 50 km/h. (a) How long does the train take to reach 85 km/h? (b) What distance does the train cover during the acceleration from 50 km/h to 85 km/h? a) To get the time I can use the given variables in the formula: v= v 0 + at v 0 = 59 km/hr = 16.4 m/s v = 85 km/hr = 23.6 m/s a = 0.68 m/s/s 23.6m/s = 16.4m/s + (0.68m/s/s)t 23.6m/s - 16.4m/s = (0.68m/s/s)t 7.2m/s = (0.68m/s/s)t (7.2m/s)/(0.68m/s/s) = t t= 10.588s b) To calculate the distance I use the formula: x = [(v 0 + v)/2]*t v 0 = 59 km/hr = 16.4 m/s v = 85 km/hr = 23.6 m/s a) time for train to reach 85 km/hr = 11s
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t = 11s x= [(16.4m/s + 23.6 m/s)/2]*11s x= [(40m/s)/2]*11s x= 20m/s * 11s x= 220 m Prob. 3.1.3 You throw a stone vertically upward at 20 m/s from a cliff that is 100 m high. (a) When does the stone return to the level at which it left your hand (presumed to be at the height of the cliff)? (b) What distance does the stone travel in reaching the ground assuming it just misses the edge of the cliff? a) I assume air resistance is ignored in this problem. I can easily solve for the time it will take to reach the top of its flight by using the velocity as zero at the highest point. The time to the highest point will be equal to the time it will take to return to its original level. As a result; doubling the time from to the maximum height will give me the total time elapsed for the stone to return to its original level. v 0 = 20 m/s v (top of flight) = 0m/s a = -9.8m/s/s v= v 0 + at 0m/s = 20m/s + (-9.8m/s/s)t -20 m/s = (-9.8m/s/s)t (-20m/s)/(-9.8m/s/s) = t b) distance covered = 220 m
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t = 2.0s Time to return to original level = 2t 2*2.0 = 4.0s b) To solve for the total distance the stone travels to the bottom of the cliff, I need to solve for the distance to the maximum height, double it, and add that to the height of the cliff. I have already solved for the time to reach the maximum height, and I can use that to calculate the distance from the top of the cliff to the top of its flight. v
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phhww3 - Physics 590 Homework, Week 3 Week 3, Homework 1...

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