HW2 - homework 02 GAUTHIER, JOSEPH Due: Sep 14 2007, 1:00...

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Unformatted text preview: homework 02 GAUTHIER, JOSEPH Due: Sep 14 2007, 1:00 am 1 Question 1, chap 2, sect 1. part 1 of 1 10 points The average speed of an orbiting space shuttle is 20000 mi / h. The shuttle is orbit- ing about 261 mi above the Earths surface. Assume the Earths radius is 3963 mi. How long does it take to circle the earth? Correct answer: 1 . 32701 h (tolerance 1 %). Explanation: The distance traveled by the space shuttle in one orbit is 2 (radius of the orbit) = 2 ( R e + h ) Hence, the required time is t = 2 ( R e + h ) v s = 2 (3963 mi + 261 mi) 20000 mi / h = 1 . 32701 h Question 2, chap 2, sect 1. part 1 of 1 10 points You plan a trip on which you want to aver- age a speed of 87 km / h. You cover the first half of the distance at an average speed of only 63 km / h. What must be your average speed in the second half of the trip to meet your goal? Correct answer: 140 . 538 km / h (tolerance 1 %). 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 = (87 km / h)(63 km / h) 2(63 km / h)- (87 km / h) = 140 . 538 km / h . Question 3, 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. None of these graphs are correct. 2. t x t Q t R t S t P 3. t x t Q t R t S t P correct homework 02 GAUTHIER, JOSEPH Due: Sep 14 2007, 1:00 am 2 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 7. t x t Q t R t S t P 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 4, chap 2, sect 2. part 1 of 1 10 points An oceanic depth-sounding vessel surveys the ocean bottom with ultrasonic waves that travel 1530 m / s in seawater....
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HW2 - homework 02 GAUTHIER, JOSEPH Due: Sep 14 2007, 1:00...

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