# HW2 - homework 02 GAUTHIER JOSEPH Due 1:00 am 1 Question 1...

This preview shows pages 1–3. Sign up to view the full content.

This preview has intentionally blurred sections. Sign up to view the full version.

View Full Document
This is the end of the preview. Sign up to access the rest of the document.

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....
View Full Document

## This note was uploaded on 03/19/2008 for the course M 408M taught by Professor Gilbert during the Fall '07 term at University of Texas.

### Page1 / 9

HW2 - homework 02 GAUTHIER JOSEPH Due 1:00 am 1 Question 1...

This preview shows document pages 1 - 3. Sign up to view the full document.

View Full Document
Ask a homework question - tutors are online