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02 homework GAUTHIER, JOSEPH Due: Sep 14 2007, 1:00 am For the entire trip, 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 orbiting about 261 mi above the Earth's surface. Assume the Earth's 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(Re + h) Hence, the required time is t= 2(Re + h) vs 2(3963 mi + 261 mi) = 20000 mi/h = 1.32701 h t = t1 + t2 d d d = + v 2 v1 2 v2 Multiplying by the LCD of (2 v v1 v2 ) yields 2v1 v2 d = v v2 d + v v1 d 2 v1 v2 = v v2 + v v1 2 v1 v2 - v v2 = v v1 Solving for v2 we have v2 = v v1 2 v1 - v (87 km/h)(63 km/h) = 2(63 km/h) - (87 km/h) = 140.538 km/h . 1 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. v t tQ tR tS tP 0 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. x 2. 0 x 3. t 0 correct tQ tR tS tP tQ tR tS tP t Question 2, chap 2, sect 1. part 1 of 1 10 points You plan a trip on which you want to average 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 d half of the trip will cover a distance of . 2 For the first half of the trip, t1 = d d1 = = v1 v1 2 v1 1 2 d For the second half of the trip, t2 = d d d2 = = v2 v2 2 v2 1 2 homework 02 GAUTHIER, JOSEPH Due: Sep 14 2007, 1:00 am x 4. t 0 x 5. t 0 6. x 0 x 7. 0 x 8. 0 9. x tQ tR tS tP t t tQ tR tS tP tQ tR tS tP t tQ tR tS tP Let : t v = 1530 m/s 6s =3s t= 2 tQ tR tS tP Question 4, chap 2, sect 2. part 1 of 1 10 points 2 An oceanic depth-sounding vessel surveys the ocean bottom with ultrasonic waves that travel 1530 m/s in seawater. How deep is the water directly below the vessel if the time delay of the echo to the ocean floor and back is 6 s? Correct answer: 4590 m (tolerance 1 %). Explanation: d = v t = (1530 m/s)(3 s) = 4590 m . Question 5, chap 2, sect 3. part 1 of 1 10 points A particle moves in a straight line with velocity v(t) = 3 t2 - 6 t. If it initially starts moving from 0 (where x = 0), then its position x(t) is equal to 1. x(t) = t3 - 3 t2 + 1 2. x(t) = 6 t - 6 3. x(t) = 6 t - 3 t2 4. x(t) = 6 t 5. x(t) = t3 - 3 t2 correct tQ tR tS tP 0 Explanation: The displacement is the integral of the velocity with respect to time: x= v dt . Because the velocity increases linearly from zero at first, then remains constant, then decreases linearly to zero, the displacement will increase at first proportional to time squared, then increase linearly, and then increase proportional to negative time squared. From these facts, we can obtain the correct answer. x t 0 tQ tR tS tP Explanation: v(t) = 3 t2 - 6 t ds(t) = 3 t2 - 6 t dt ds(t) dt = 3 t2 - 6 t dt dt s(t) = t3 - 3 t2 + C homework 02 GAUTHIER, JOSEPH Due: Sep 14 2007, 1:00 am x(0) = 0 0 = C, so s(t) = t3 - 3 t2 . Question 6, chap 2, sect 3. part 1 of 5 10 points Consider the displacement curve: OABC Displacement vs Time 2 1 x (m) 0 -1 -2 0 1 B O C A 3 axis. The key here is to know that velocity uses net displacement in its equation while speed uses total distance traveled. v OA = xA - xO tA - tO 2-0 = 1-0 = +2 m/s . The values for the average velocity come directly from evaluating the definitions. Question 7, chap 2, sect 3. part 2 of 5 10 points Choose the appropriate quantity for the average velocity v OB for the motion from point O to point B. 1. v OB = +2 m/s 2. v OB = - 3 m/s 3. v OB = + 3 m/s 4. v OB = -2 m/s 5. v OB = 0 m/s correct Explanation: xB - xO tB - tO 0-0 = = 0 m/s . 2-0 2 t (s) Choose the appropriate quantity for the average velocity v OA from point O to A. 1. v OA = - 3 m/s 2. v OA = 0 m/s 3. v OA = -2 m/s 4. v OA = +2 m/s correct 5. v OA = + 3 m/s Explanation: Definitions: displacement Average velocity: v = time distance Average speed: s = time dx Instantaneous velocity: v = dt Instantaneous speed: s = |v| Solution: Note that the displacement x is on the vertical axis and that time t is on the horizontal v OB = Question 8, chap 2, sect 3. part 3 of 5 10 points Choose the appropriate quantity for the average speed sOB for the motion from point O to point B. 1. sOB = +2 m/s correct 2. sOB = + 3 m/s 3. sOB = 0 m/s homework 02 GAUTHIER, JOSEPH Due: Sep 14 2007, 1:00 am 4. sOB = -2 m/s 5. sOB = - 3 m/s 4 Explanation: sOB = Choose the appropriate quantity for the instantaneous speed sB at point B. 1. sB = - 3 m/s 2. sB = 0 3. sB = -2 m/s 4. sB = +2 m/s correct 5. sB = + 3 m/s Explanation: Instantaneous speed, is simply the absolute value of instantaneous velocity because speed has magnitude but no direction. sB = |vB | = | - 2| = +2 m/s . Question 11, chap 2, sect 4. part 1 of 1 10 points A car traveling in a straight line has a velocity of 2.38 m/s at some instant. After 5.46 s, its velocity is 8.31 m/s. What is its average acceleration in this time interval? Correct answer: 1.08608 m/s2 (tolerance 1 %). Explanation: The average acceleration is aav = v t vf - vi = t 8.31 m/s - 2.38 m/s = 5.46 s = 1.08608 m/s2 |xA - xO | + |xB - xA | tB - tO 2+2 = = +2 m/s . 2-0 The values for the average speed come directly from evaluating the definitions. Question 9, chap 2, sect 3. part 4 of 5 10 points Choose the appropriate quantity for the instantaneous velocity vB at point B. 1. vB = -2 m/s correct 2. vB = - 3 m/s 3. vB = +2 m/s 4. vB = 0 m/s 5. vB = + 3 m/s Explanation: The instantaneous velocity at point B can be obtained by first finding an expression for the graph describing the position of the moving object and taking its derivative evaluated at time tB . However, since the graph near B is linear, it is simpler to calculate the slope of the line over the interval from A to B. The result correctly describes the instantaneous velocity at B because the derivative of a straight line is constant at all points on the line and can be obtained for our case by x - xA vB = B tB - tA 0-2 = -2 m/s . = 2-1 Question 10, chap 2, sect 3. part 5 of 5 10 points Question 12, chap 2, sect 4. part 1 of 1 10 points A car going north on Guadalupe approaches a red light at 24th street. The driver homework 02 GAUTHIER, JOSEPH Due: Sep 14 2007, 1:00 am applies the brakes. Which of the following is then true? 1. a < 0, v > 0 correct 2. a = 0, v > 0 3. a > 0, v < 0 4. a = 0, v < 0 5. a < 0, v < 0 6. a > 0, v > 0 Explanation: The speed is decreasing, so the acceleration is negative. But the car is still moving forward until it stops, so the velocity is positive. Question 13, chap 2, sect 4. part 1 of 1 10 points What is the acceleration of a vehicle that changes its velocity from 500 km/h to a dead stop in 20 s ? Correct answer: -6.94444 m/s2 (tolerance 1 %). Explanation: The acceleration is v a= t 1h 0 - 500 km/h 1000 m = 20 s 1 km 60 min 1 min 60 s = -6.94444 m/s2 . Question 14, chap 2, sect part 5. 1 of 1 10 points An electron, starting from rest and moving with a constant acceleration, travels 1.8 cm in 12 ms. What is the magnitude of this acceleration in km/s2 ? Correct answer: 0.25 km/s2 (tolerance 1 %). Explanation: Question 15, chap 2, sect 5. part 1 of 2 10 points 5 An electron starting from rest with constant acceleration will travel a distance of 1 d = at2 2 in t seconds. Given the time and distance traveled we solve for a: a= 2d t2 making any necessary unit conversions along the way. Two particles moving along parallel paths in the same direction pass the same point at the same time. Particle A has an initial velocity of 8 m/s and an acceleration of 4.7 m/s2 . Particle B has an initial velocity of 4.5 m/s and an acceleration of 6.2 m/s2 . a) At what time will B pass A? Correct answer: 4.66667 s (tolerance 1 %). Explanation: Basic Concept: 1 s = v0 t + at2 2 for each particle. Solution: Particle B will pass particle A when the positions are the same, so equate the positions and solve the resulting equation for t: 1 1 v0A t + aA t2 = v0B t + aB t2 2 2 2v0A t + aA t2 = 2v0B t + aB t2 aA t2 - aB t2 + 2v0A t - 2v0B t = 0 t(aA t - aB t + 2v0A - 2v0B ) = 0 t (aA - aB )t + 2(v0A - v0B ) = 0 t= 2(v0B - v0A ) (aA - aB ) homework 02 GAUTHIER, JOSEPH Due: Sep 14 2007, 1:00 am The trivial solution t = 0 must be rejected. tAB + tBC + tCD = 4.20 min vA = vD = 0 m/s Question 16, chap 2, sect 5. part 2 of 2 10 points b) At what position will the faster particle pass the slower? Correct answer: 88.5111 m (tolerance 1 %). Explanation: the time found in Part 1 into the following position equation: 1 1 s = v0A t + aA t2 = v0B t + aB t2 . 2 2 Question 17, chap 2, sect 4. part 1 of 3 10 points A train travels between stations 1 and 2, as shown in the figure. The engineer of the train is instructed to start from rest at station 1 and accelerate uniformly between points A and B, then coast with a uniform velocity between points B and C, and finally accelerate uniformly between points C and D until the train stops at station 2. The distances AB, BC, and CD are all equal, and it takes 4.20 min to travel between the two stations. Assume that the uniform accelerations have the same magnitude, even when they are opposite in direction. Station A Station B Solution: t = x vavg 6 Between points A and B, tAB = xAB 2xAB xAB = vB +0 = vABavg vB 2 Because the train's velocity is constant from B to C, tBC = xAB xBC = vB vB Between points C and D, tCD = Thus 2xAB xAB 2xAB + + = 4.2 min vB vB vB 5xAB = 4.2 min vB 4.2 min xAB = vB 5 = 0.84 min and tAB = 2xAB = 1.68 min vB xCD xAB 2xAB = 0+vB = vCDavg vB 2 A B C D Question 18, chap 2, sect 4. part 2 of 3 10 points b) How much of this 4.20 min period does the train spend between points B and C? Correct answer: 0.84 min (tolerance 1 %). Explanation: Solution: tBC = xAB = 0.84 min vB Explanation: Basic Concept: a) How much of this 4.20 min period does the train spend between points A and B? Correct answer: 1.68 min (tolerance 1 %). vavg = Given: x t xAB = xBC = xCD Question 19, chap 2, sect 4. part 3 of 3 10 points homework 02 GAUTHIER, JOSEPH Due: Sep 14 2007, 1:00 am c) How much of this 4.20 min period does the train spend between points C and D? Correct answer: 1.68 min (tolerance 1 %). 7 1 containing v, a, and t . Choose the positive direction to be up; then a = -g and 0 = v0 + (-g) tup or Explanation: Solution: tCD = 2xAB = 1.68 min vB tup = Question 20, chap 2, sect 6. part 1 of 1 10 points A ball is thrown upward. Its initial vertical speed is 11.4 m/s , acceleration of gravity is 9.8 m/s2 , and maximum height hmax are shown in the figure below. Neglect: Air resistance. The acceleration of gravity is 9.8 m/s2 . v0 g (11.4 m/s) = (9.8 m/s2 ) = 1.16327 s . Question 21, chap 2, sect 6. part 1 of 2 10 points A ball has an initial speed of 15 m/s. The acceleration of gravity is 9.8 m/s2 . 2.4 s t hmax 15 m/s 9.8 m/s2 11.4 m/s h y What will be its position after 2.4 s if it is thrown a) down with an initial speed of 15 m/s? Correct answer: -64.224 m (tolerance 1 %). Explanation: Basic Concepts: This problem involves initial velocities directed both downward and upward, so all downward motion must be negative. Assuming the initial position is ho = 0, the position at any time t is given by hf = vo t - (1) 1 2 gt . 2 What is its time interval, tup , between the release of the ball and the time it reaches its maximum height? Correct answer: 1.16327 s (tolerance 1 %). Explanation: Let : v0 = 11.4 m/s and g = 9.8 m/s2 . Basic Concept: For constant acceleration, we have v = v0 + a t . Solution: The velocity at the top is zero. Since we know velocities and acceleration, Eq. Let : t = 2.4 s , vo = 15 m/s , and g = 9.8 m/s2 . homework 02 GAUTHIER, JOSEPH Due: Sep 14 2007, 1:00 am Solution: h1 = -v t - 1 2 gt 2 = -(15 m/s) (2.4 s) 1 - (9.8 m/s2 ) (2.4 s)2 2 = -64.224 m . 8 Given: For the first time interval, (while the engines are running), vi,1 = 45.8 m/s a1 = 2.49 m/s2 y1 = 150 m For the second time interval (after the engines stop), vi,2 = vf,1 vf,2 = 0 m/s a2 = -9.81 m/s2 Solution: Question 22, chap 2, sect 6. part 2 of 2 10 points y h 15 m/s t 2.4 s b) up with an initial speed of 15 m/s? Correct answer: 7.776 m (tolerance 1 %). 1 h2 = v t - g t2 2 = (15 m/s) (2.4 s) 1 - (9.8 m/s2 ) (2.4 s)2 2 = 7.776 m . Question 23, chap 2, sect 6. part 1 of 3 10 points A model rocket is launched straight upward with an initial speed of 45.8 m/s. It accelerates with a constant upward acceleration of 2.49 m/s2 until its engines stop at an altitude of 150 m. a) What is the maximum height reached by the rocket? Correct answer: 294.987 m (tolerance 1 %). Explanation: Basic Concept: 2 2 vf = vi + 2ay 2 vf,1 = (45.8 m/s)2 + 2 2.49 m/s2 (150 m) = 2844.64 m2 /s2 vf,1 = 2844.64 m2 /s2 = 53.3352 m/s which is +53.3352 m/s since the motion is upward. This is the initial velocity for the second time interval, which has a final velocity of zero, so the equation simplifies to y2 = 2 -vi,2 Explanation: 2a2 -(53.3352 m/s)2 = 2 (-9.81 m/s2 ) = 144.987 m and the maximum height is ymax = y1 + y2 = 150 m + 144.987 m = 294.987 m Question 24, chap 2, sect 6. part 2 of 3 10 points b) When does the rocket reach maximum height? Correct answer: 8.46299 s (tolerance 1 %). homework 02 GAUTHIER, JOSEPH Due: Sep 14 2007, 1:00 am Explanation: Basic Concept: y = vavg t = vi + vf t 2 Solution: tdown = = 2y a 9 Solution: For the first time interval, tup,1 = 2y1 vi,1 + vf,1 2(150 m) = 45.8 m/s + 53.3352 m/s = 3.02617 s 2(-294.987 m) -9.81 m/s2 = 7.755 s Thus the total time is ttot = tup + tdown = 8.46299 s + 7.755 s = 16.218 s For the second time interval, vi,2 = vf,1 and vf,2 = 0 m/s, so tup,2 = 2y2 vi,2 2(144.987 m) = 53.3352 m/s = 5.43682 s Thus the total time up is ttot,up = tup,1 + tup,2 = 3.02617 s + 5.43682 s = 8.46299 s Question 25, chap 2, sect 6. part 3 of 3 10 points c) How long is the rocket in the air? Correct answer: 16.218 s (tolerance 1 %). Explanation: Basic Concept: For the trip down, the equation simplifies to 1 y = a(t)2 2 since vi = 0 m/s. Given: For the trip down, y = -294.987 m vi,down = 0 m/s a2 = -9.81 m/s2

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