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Unformatted text preview: Question 1, chap 2, sect 2. part 1 of 1 10 points The graph below shows the velocity v as a function of time t for an object moving in a straight line. t v t Q t R t S t P Which of the following graphs shows the corresponding displacement x as a function of time t for the same time interval? 1. t x t Q t R t S t P 2. t x t Q t R t S t P 3. t x t Q t R t S t P 4. t x t Q t R t S t P 5. t x t Q t R t S t P 6. t x t Q t R t S t P correct 7. None of these graphs are correct. 8. t x t Q t R t S t P 9. t x t Q t R t S t P Explanation: The displacement is the integral of the ve locity with respect to time: vectorx = integraldisplay vectorv dt . Because the velocity increases linearly from zero at first, then remains constant, then de creases linearly to zero, the displacement will increase at first proportional to time squared, then increase linearly, and then increase pro portional to negative time squared. From these facts, we can obtain the correct answer. t x t Q t R t S t P Question 2, chap 2, sect 2. part 1 of 2 10 points You drive a car 9 h at 46 km / h, then 9 h at 66 km / h. What is your average velocity? Correct answer: 56 km / h (tolerance 1 %). Explanation: Basic Concept Average velocity is defined by v av = d t Solution If you drive for t hours at a speed of v 1 and for the same time at a speed of v 2 , then d = d 1 + d 2 = v 1 t + v 2 t in a time of t = t 1 + t 2 = 2 t so that the average velocity is v av = v 1 t + v 2 t 2 t = v 1 + v 2 2 = (46 km / h) + (66 km / h) 2 = 56 km / h . Question 3, chap 2, sect 2. part 2 of 2 10 points What is your average velocity if you drive a distance of 504 km at a speed of 46 km / h, then the same distance at a speed of 66 km / h? Correct answer: 54 . 2143 km / h (tolerance 1 %). Explanation: If you drive the distance d at v 1 and the same distance at v 2 , then d = d 1 + d 2 = 2 d and t = t 1 + t 2 = d v 1 + d v 2 so that the average velocity will be v av = 2 d d v 1 + d v 2 v 1 v 2 v 1 v 2 = 2 d v 1 v 2 d v 2 + d v 1 = 2 v 1 v 2 v 1 + v 2 = 2 (46 km / h) (66 km / h) (46 km / h) + (66 km / h) = 54 . 2143 km / h . Question 4, chap 2, sect 2. part 1 of 2 10 points A particle moves according to the equation x = (10 m / s 2 ) t 2 where x is in meters and t is in seconds. Find the average velocity for the time in terval from t 1 = 2 . 16 s to t 2 = 4 . 63 s. Correct answer: 67 . 9 m / s (tolerance 1 %). Explanation: x = (10 m / s 2 ) t 2 v ave = x t = (10 m / s 2 ) ( t 2 2 t 2 1 ) t 2 t 1 = (10 m / s 2 ) [(4 . 63 s) 2 (2 . 16 s) 2 ] (4 . 63 s) (2 . 16 s) = 67 . 9 m / s Question 5, chap 2, sect 2. part 2 of 2 10 points Find the average velocity for the time in terval from t 1 = 2 . 16 s to t 3 = 2 . 26 s....
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This note was uploaded on 12/04/2008 for the course PHY 302K taught by Professor Kaplunovsky during the Fall '08 term at University of Texas at Austin.
 Fall '08
 Kaplunovsky

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