Practice Homework 2 Solutions

# Practice Homework 2 Solutions - practice 02 ALIBHAI ZAHID...

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practice 02 – ALIBHAI, ZAHID – Due: Jan 28 2007, 4:00 am 1 Question 1 part 1 of 2 10 points A person travels by car from one city to another. She drives for 31 . 2 min at 69 . 8 km / h, 8 . 5 min at 112 km / h, 47 . 4 min at 42 . 3 km / h, and spends 16 . 5 min along the way eating lunch and buying gas. Determine the distance between the cities along this route. Correct answer: 85 . 5797 km (tolerance ± 1 %). Explanation: Distances traveled are x 1 = v 1 t 1 x 2 = v 2 t 2 x 3 = v 3 t 3 Then the total distance traveled is x = x 1 + x 2 + x 3 = (36 . 296 km) + (15 . 8667 km) + (33 . 417 km) = 85 . 5797 km . Question 2 part 2 of 2 10 points Determine the average speed for the trip. Correct answer: 49 . 5635 km / h (tolerance ± 1 %). Explanation: And, time spent is t = t 1 + t 2 + t 3 + t other = (31 . 2 min) + (8 . 5 min) + (47 . 4 min) + (16 . 5 min) = 1 . 72667 h . Hence v av = x t = 85 . 5797 km 1 . 72667 h = 49 . 5635 km / h . Question 3 part 1 of 2 10 points A runner is jogging at a steady v r = 5 . 1 km / hr. When the runner is L = 6 . 8 km from the finish line a bird begins flying from the runner to the finish line at v b = 30 . 6 km / hr (6 times as fast as the runner). When the bird reaches the finish line, it turns around and flies back to the runner. Even though the bird is a dodo, we will assume that it occupies only one point in space, i.e. , a zero length bird. L v b v r finish line How far does the bird travel? Correct answer: 11 . 6571 km (tolerance ± 1 %). Explanation: Let, dodo birds fly, and d r be the distance the runner travels. d b be the distance the bird travels. v r be the speed of the runner. v b be the speed of the bird. L = d r be the original distance to the finish line. L 1 be the distance to the finish line after the first encounter. . . . L i be the distance to the finish line after the i th encounter. finish line L 1 d r 1 d b 1 Since the bird travels 6 times as fast as the runner at the first meeting between the bird and runner, d b 1 = 6 d r 1 . (1)

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practice 02 – ALIBHAI, ZAHID – Due: Jan 28 2007, 4:00 am 2 The sum of the bird’s and runner’s distances is 6 times L . d b 1 + d r 1 = 2 L . (2) Therefore, substituting for d b 1 from Eq. (1) d r 1 + 6 d r 1 = 2 L d r 1 = 2 7 L = 2 7 (6 . 8 km) = 1 . 94286 km . (3) Thus the distance the bird flies is d b 1 = 6 d r 1 = 12 7 L = 12 7 (6 . 8 km) = 11 . 6571 km , (4) and the distance for the runner to travel after this first encounter is L 1 = 5 7 L = 5 7 (6 . 8 km) = 4 . 85714 km . Question 4 part 2 of 2 10 points After this first encounter, the bird then turns around and flies from the runner back to the finish line, turns around again and flies back to the runner. The bird repeats the back and forth trips until the runner reaches the finish line. How far does the bird travel from the be- ginning? ( i.e. , include the distance traveled to the first encounter) Correct answer: 40 . 8 km (tolerance ± 1 %). Explanation: Repeating this scenario a second time the distance for the runner to travel after the second encounter is L 2 = 5 7 L 1 = parenleftbigg 5 7 parenrightbigg 2 L , and the third time L 3 = 5 7 L 2 = parenleftbigg 5 7 parenrightbigg 3 L , and the i th time L i = 5 7 L i 1 = parenleftbigg 5 7 parenrightbigg i L . (5) Note: The distance the bird travels between the ( i 1) th and i th time is [see Eq. (4)] d b i = 12 7 L parenleftbigg 5 7
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