HW12-solutions

# HW12-solutions - lee (cyl327) HW12 gentle (56245) 1 This...

This preview shows pages 1–3. Sign up to view the full content.

This preview has intentionally blurred sections. Sign up to view the full version.

View Full Document
This is the end of the preview. Sign up to access the rest of the document.

Unformatted text preview: lee (cyl327) HW12 gentle (56245) 1 This print-out should have 50 questions. Multiple-choice questions may continue on the next column or page find all choices before answering. This is a set of review prolems. 001 (part 1 of 3) 10.0 points A particle moving uniformly along the x axis is located at 9 . 18 m at 1 . 6 s and at 3 . 19 m at 3 . 54 s. Find its displacement during this time in- terval. Correct answer: 5 . 99 m. Explanation: Let : x f = 3 . 19 m and x i = 9 . 18 m . The displacement is x = x f x i = 3 . 19 m 9 . 18 m = 5 . 99 m . 002 (part 2 of 3) 10.0 points What is its average velocity during this time interval? Correct answer: 3 . 08763 m / s. Explanation: Let : t i = 1 . 6 s and t f = 3 . 54 s . The average velocity is v = x t = x f x i t f t i = 3 . 19 m 9 . 18 m 3 . 54 s 1 . 6 s = 3 . 08763 m / s . 003 (part 3 of 3) 10.0 points Calculate the particles average speed during this time interval. Correct answer: 3 . 08763 m / s. Explanation: The average speed is the magnitude of the average velocity: speed avg = | v | = | 3 . 08763 m / s | = 3 . 08763 m / s . 004 10.0 points A flowerpot falls from the ledge of an apart- ment building. A person in an apartment below, coincidentally holding a stopwatch, no- tices that it takes 0 . 6 s for the pot to fall past his window, which is 14 m high. How far above the top of the window is the ledge from which the pot fell? The accelera- tion of gravity is 9 . 81 m / s 2 . Correct answer: 21 . 1909 m. Explanation: Let : t win = 0 . 6 s , h win = 14 m , and g = 9 . 81 m / s 2 . Using a constant-acceleration equation, ex- press the distance y below the ledge from which the pot fell as a function of time: y = y + v + 1 2 g t 2 = 1 2 g t 2 since v = y = 0. The distance to the top is y top = 1 2 g t 2 top and to the bottom y bottom = 1 2 g ( t top + t win ) 2 , where t win = t bottom t top . y win = 1 2 g [( t top + t win ) 2 t 2 top ] = 1 2 g (2 t top t win ) + 1 2 g ( t win ) 2 t top = y win g t win 1 2 t win = 14 m (9 . 81 m / s 2 ) (0 . 6 s) 1 2 (0 . 6 s) = 2 . 07853 s and lee (cyl327) HW12 gentle (56245) 2 y top = 1 2 (9 . 81 m / s 2 ) (2 . 07853 s) 2 = 21 . 1909 m . 005 (part 1 of 2) 10.0 points The coefficient of static friction between a rubber tire and the road surface is 0 . 55. The acceleration of gravity is 9 . 81 m / s 2 . What is the maximum acceleration of a 950 kg four-wheel-drive truck if the road makes an angle of 14 with the horizontal and the truck is climbing? Correct answer: 2 . 86198 m / s 2 . Explanation: Let : s = 0 . 55 , m = 950 kg , and = 14 ....
View Full Document

## This note was uploaded on 02/28/2012 for the course PHY 316 taught by Professor Markert during the Fall '10 term at University of Texas at Austin.

### Page1 / 18

HW12-solutions - lee (cyl327) HW12 gentle (56245) 1 This...

This preview shows document pages 1 - 3. Sign up to view the full document.

View Full Document
Ask a homework question - tutors are online