Homework #2 solution

Homework #2 solution - 2 2 2 2 2 2 2 2 ( 0.03) 18.5 ; 2 (...

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Fall 2008 A. G. Hobeika CEE3604--Homework #2 Solution Question 1 (Question 2.28 in the textbook) An Engineering student claims that a country road can be safely negotiated at 70mi/h in rainy weather. Because of the winding nature of the road, one stretch of level pavement has a sight distance of only 590ft. Assuming practical stopping distance, comment on the student’s claim
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Question 2 A student trying to test the braking ability of her car determined that she needed 18.5 ft more to stop her car when driving downhill on a road segment of 5 percent grade than when driving downhill at the same speed along another segment of 3 percent grade. Determine the speed at which the student conducted her test and the braking distance on the 5 percent grade if the student is traveling at the test speed in the uphill direction. 2 2 2 Let = downhill braking distance on 5 percent grade ( 18.5) downhill braking distance on 3 percent grade downhill with 5% grade : 2 ( ) 2 ( 0.05) downhill with 3% grade : 18.5 2 b x x ft V V x D a a g G g g g V x - = = = = - - - =
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Unformatted text preview: 2 2 2 2 2 2 2 2 ( 0.03) 18.5 ; 2 ( 0.05) 2 ( 0.03) 11.2 /sec , 32.2 /sec 0.111 0.104 18.5 75 /sec Determine braking distance on uphill grade 75 220 11.2 2 ( ) 2(32.2)( 0.05) 32.2 b a g g V V a a g g g g a ft g ft V V V ft V D ft a g G g- =---= =-= = = = = + + Q Question 3 Calculate the minimum passing sight distance required for a two-lane rural roadway that has a posted speed limit of 45 mi/h. The local traffic engineer conducted a speed study of the subject road and found the following: average speed of the passing vehicle was 47 mi/ h with an average acceleration of 1.43 mi/h/sec, and average speed of impeder vehicles was 40mi/h. 1 2 1 1 1 2 2 2 3 4( ) and 1.43, =10(s) are obtained in the Table in slide 6 of the PPT 1.43(4.0) 1.47 ( ) 1.47(4.0)(47 7 ) 252 2 2 The distance is obtained from 1.47 1.47(47)(10) 690 =180ft t s a t at d t u m ft d d ut ft d = = =-+ =- + = = = = 4 2 1 2 3 4 (from the same Table) = 2/3 460.6 1583 d d ft total d d d d ft = = + + + =...
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Homework #2 solution - 2 2 2 2 2 2 2 2 ( 0.03) 18.5 ; 2 (...

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