This preview shows pages 1–2. Sign up to view the full content.
This preview has intentionally blurred sections. Sign up to view the full version.
View Full Document
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....
View Full
Document
 Spring '07
 LECLAIR,A
 mechanics, Work

Click to edit the document details