p201f04-homework03-sol

# p201f04-homework03-sol - Version One – Homework 3 –...

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Unformatted text preview: Version One – Homework 3 – Heinz – 81204 – Sep 09, 2004 1 This print-out should have 14 questions. Multiple-choice questions may continue on the next column or page – find all choices before answering. The due time is Central time. Paths Cross 02 02:02, trigonometry, numeric, > 1 min, nor- mal. 001 (part 1 of 1) 10 points Runner A is initially 4 mi West of a flag- pole and is running with a constant velocity of 6 mi / h due East. Runner B is initially 3 mi East of the flagpole and is running with a con- stant velocity of 5 mi / h due West. Consider East to be the positive direction. What is the displacement of runner B from the flagpole when their paths cross? Correct answer:- . 181818 mi. Explanation: Let the flagpole be the origin. The equa- tions of position as a function of time for runners A and B, respectively, are x A ( t ) = x A + v A t (1) x B ( t ) = x B + v B t (2) Their paths cross when x A ( t ) = x B ( t ); or x A + v A t = x B + v B t Solving for t yields the time when they cross. t = + x A- x B- v A + v B = (- 4 mi)- (3 mi)- (6 mi / h) + (- 5 mi / h) = 0 . 636364 h . Hence, from (2) the position of runner B at t = 0 . 636364 h is x B f = x B + v B t = (3 mi) + (- 5 mi / h)(0 . 636364 h) =- . 181818 mi Plan a Trip 02:02, trigonometry, numeric, > 1 min, nor- mal. 002 (part 1 of 1) 10 points You plan a trip on which you want to average a speed of 90 km / h. You cover the first half of the distance at an average speed of only 48 km / h. What must be your average speed in the second half of the trip to meet your goal? Correct answer: 720 km / h. Explanation: Basic Concepts If d is the total distance covered, then each half of the trip will cover a distance of d 2 . For the first half of the trip, t 1 = d 1 v 1 = 1 2 d v 1 = d 2 v 1 For the second half of the trip, t 2 = d 2 v 2 = 1 2 d v 2 = d 2 v 2 For the entire trip, t = t 1 + t 2 d v = d 2 v 1 + d 2 v 2 Multiplying by the LCD of (2 v v 1 v 2 ) yields 2 v 1 v 2 d = v v 2 d + v v 1 d 2 v 1 v 2 = v v 2 + v v 1 2 v 1 v 2- v v 2 = v v 1 Solving for v 2 we have v 2 = v v 1 2 v 1- v = (90 km / h)(48 km / h) 2(48 km / h)- (90 km / h) = 720 km / h . Bullet Acceleration 02:04, trigonometry, numeric, > 1 min, nor- mal....
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## This note was uploaded on 07/02/2011 for the course PHYS 201 taught by Professor Woodahl during the Fall '09 term at Indiana.

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p201f04-homework03-sol - Version One – Homework 3 –...

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