solution-hw3

# solution-hw3 - lai(yl8859 – HW03 – Ross –(15200 1...

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Unformatted text preview: lai (yl8859) – HW03 – Ross – (15200) 1 This print-out should have 8 questions. Multiple-choice questions may continue on the next column or page – find all choices before answering. 001 10.0 points A train is moving parallel and adjacent to a highway with a constant speed of 23 m / s. Ini- tially a car is 59 m behind the train, traveling in the same direction as the train at 35 m / s and accelerating at 4 m / s 2 . What is the speed of the car just as it passes the train? Correct answer: 47 . 8193 m / s. Explanation: Let : x ot = 59 m , v t = 23 m / s , a t = 0 m / s 2 , x oc = 0 m , v oc = 35 m / s , and a c = 4 m / s 2 . After a time t , the car will be at position x c = v oc t + 1 2 a c t 2 , and the train at x t = x ot + v t t . When the car catches up with the train after time t , x c = x t v oc t + 1 2 a t 2 = x ot + v t t a t 2 + 2 ( v oc- v t ) t- 2 x ot = 0 Applying the quadratic equation t =- b ± √ b 2- 4 a c 2 a . b = 2 ( v oc- v t ) t = 2 (35 m / s- 23 m / s) (3 . 20484 s) = 24 m and the discriminant b 2 + 8 a x ot = (24 m) 2 + 8 (4 m / s 2 ) (59 m) = 2464 m 2 , so t =- 24 m ± √ 2464 m 2 2 (4 m / s 2 )...
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## This note was uploaded on 02/24/2011 for the course PHYS 152 taught by Professor Button during the Winter '08 term at IUPUI.

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solution-hw3 - lai(yl8859 – HW03 – Ross –(15200 1...

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