HW_1_sol

# HW_1_sol - negligible the rider covers a total distance 2D...

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P112 Spring 2007 Homework #1 Solutions (of boxed problems only) 3. Total distance is 20 miles, total time is 1.5 h. Average speed is 13.3 miles per hour. 5. Plot of f(x) in the range of 0<x<5: 7. a) One can “guard” about 10 parking spaces b) A typical shopping trip takes about 50 minutes. Based on those estimates, one will have to wait for about 5 minutes. 9. (a) The times spent going up and down the hill are: t 1 = D/v 1 and t 2 = D/v 2 The total time is: t 1 + t 2 = D/v 1 + D/v 2 The average velocity is the total distance 2D divided by the total time: v av = 2D/( D/v 1 + D/v 2 ) = 2v 1 v 2 v 1 + v 2 As long as the distance is the same for going up and down the average velocity does not depend on D. (b) v av = 2v 1 2 /(2v 1 ) = v 1 . This makes sense because the average velocity is the same as the (constant) instantaneous velocity. (c) v av 2v 1 v 2 /(v 2 ) = 2v 1 . This also makes sense. Since the time to cover the second distance D is

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Unformatted text preview: negligible, the rider covers a total distance 2D in roughly the time it took to go the first D. Therefore, the average velocity is twice v 1 . (d) v av =2v 1 x/(x+1). If x becomes very large, the expression x/(x+1) is approximately 1 (see graph in Question 5), and v av =2v 1 , as in (c) f(x) x (e) v av ≈ 2v 1 v 2 /(v 1 ) = 2v 2 This makes sense for reasons similar to part (c). (f) See graph at right. For v 2-> ∞ , v av -> 2v 1 . (g) v av = 2(4.0 mi/h)(12 mi/h) 4.0 mi/h + 12 mi/h = 6.0 mi/h (h) Solving for v 2 from part (a): v 2 = v 1 v av 2v 1- v av Computing v 2 values: v 2 = (4.0 mi/h)(2.0 mi/h) 2(4.0 mi/h) - 2.0 mi/h = 1.33 mi/h v 2 = (4.0 mi/h)(4.0 mi/h) 2(4.0 mi/h) - 4.0 mi/h = 4.0 mi/h v 2 = (4.0 mi/h)(8.0 mi/h) 2(4.0 mi/h) - 8.0 mi/h = ∞ Not possible since v 2 cannot = ∞ . 12.0 mi/h not possible since v av is always ≤ 2v 1 = 8.0 mi/h....
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HW_1_sol - negligible the rider covers a total distance 2D...

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