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Unformatted text preview: Muraj, Hamza Homework 4 Due: Jan 30 2006, 4:00 am Inst: Florin 1 This printout should have 12 questions. Multiplechoice questions may continue on the next column or page find all choices before answering. The due time is Central time. 001 (part 1 of 1) 10 points A car traveling at a constant speed of 44 . 7 m / s passes a trooper hidden behind a billboard at t = 1s. One second later, at t = 0s, the trooper sets in chase after the car with a constant acceleration of 3 . 45 m / s 2 . At what time t does the trooper overtake the speeding car? Correct answer: 26 . 8772 s. Explanation: To solve this problem algebraically, we write expressions for the position of each ve hicle as a function of time. It is convenient to choose the origin at the position of the billboard and take t = 0 as the time the trooper begins moving. At that instant, the speeding car has already traveled a distance of 44 . 7 m because it travels at a constant speed of 44 . 7 m / s. Thus, the initial position of the speeding car is 44 . 7 m. Because the car moves with constant speed, its acceleration is zero, and applying the equation x C ( t ) = x C ( t = 0) + v C t , gives x C ( t ) = 44 . 7 m + (44 . 7 m / s) t . Note that at t = 0, this expression does give the cars correct initial position x C = x = 44 . 7 m. For the trooper, who starts from the origin at t = 0, we have x = 0, v = 0, and a = 3 . 45 m / s 2 . Hence, the position of the trooper as a function of time is x T = 1 2 a T t 2 = 1 2 (3 . 45 m / s 2 ) t 2 . The trooper overtakes the car at the instant that x T = x C , or 1 2 (3 . 45 m / s 2 ) t 2 = 44 . 7 m + (44 . 7 m / s) t . This gives the quadratic equation 1 . 725 m / s 2 t 2 44 . 7 m / s t 44 . 7 m = 0 , whose positive solution is t = 26 . 8772 s . 002 (part 1 of 3) 10 points A ball is dropped from rest at point O . After falling for some time, it passes by a window of height 3 . 9 m and it does so during time 0 . 26 s. The acceleration of gravity is 9 . 8 m / s 2 . O A B 3 . 9 m x y The ball accelerates all the way down; let v A be its speed as it passes the windows top A and v B its speed as it passes the windows bottom B . How much did the ball speed up as it passed the window; i.e. , calculate v down = v B v A ? Correct answer: 2 . 548 m / s. Explanation: Let : h = 3 . 9 m , t AB = 0 . 26 s , and g = 9 . 8 m / s 2 . O A B h t AB t y Assume: Down is positive. Muraj, Hamza Homework 4 Due: Jan 30 2006, 4:00 am Inst: Florin 2 The ball falls under a constant acceleration g , so g = v t = v B v A t and the change of its velocity during time t AB is simply ~v = ~v B ~v A = g t AB , assuming the downward direction to be positive....
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This note was uploaded on 04/12/2010 for the course PHY 58195 taught by Professor Turner during the Spring '09 term at University of Texas at Austin.
 Spring '09
 Turner

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