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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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 Spring '07
 LECLAIR,A
 mechanics, Work, 5 minutes, 12 mi, 6.0 mi, 1.33 mi, 12.0 mi

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