INSTRUCTOR'S SOLUTIONS MANUAL 14TH EDITION,Document details
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IDENTIFY: Since air resistance is ignored, the egg is in free-fall and has a constant downward acceleration of magnitude 9. Apply the constant acceleration equations to the motion of the egg. d The acceleration is constant and equal to 9. IDENTIFY: We can avoid solving for the common height by considering the relation between height, time of fall, and acceleration due to gravity, and setting up a ratio involving time of fall and acceleration due to gravity. SET UP: Let g En be the acceleration due to gravity on Enceladus and let g be this quantity on earth. Let h be the common height from which the object is dropped. IDENTIFY: Since air resistance is ignored, the boulder is in free-fall and has a constant downward acceleration of magnitude 9. Apply the constant acceleration equations to the motion of the boulder. e The acceleration is 9. f The graphs are sketched in Figure 2. The time to reach the maximum height is half the total time in the air, so the answer in part d is half the answer in part c.
Also note that 2. The boulder is going upward until it reaches its maximum height and after the maximum height it is traveling downward. The constant-acceleration kinematics formulas apply. First find the initial velocity and then the final speed. EXECUTE: a 6. When the rock. EVALUATE: The final speed is greater than the initial speed because the rock accelerated on its way down below the bridge. IDENTIFY: The acceleration is not constant, so we must use calculus instead of the standard kinematics formulas. First integrate ax to find t. Now use this result to find x t. EVALUATE: The standard kinematics formulas apply only when the acceleration is constant. IDENTIFY: The acceleration is not constant, but we know how it varies with time.
EVALUATE: The time in part b is less than IDENTIFY: The acceleration is not constant so the constant acceleration equations cannot be used. SET UP:. Then, at 0. a IDENTIFY: Integrate a x t to find vx t and then integrate vx t to find x t. For earlier times a x is positive so vx is still increasing. negative and vx is decreasing. IDENTIFY: a t is the slope of the v versus t graph and the distance traveled is the area under the v versus t graph. SET UP: The v versus t graph can be approximated by the graph sketched in Figure 2. The acceleration during the first 1. We need to break up the race into segments. The total time is 6. The total time is His total time is IDENTIFY: We know the vertical position of the lander as a function of time and want to use this to find its velocity initially and just before it hits the lunar surface.
The initial velocity is IDENTIFY: The brick has a constant downward acceleration, so we can use the usual kinematics formulas. We know that it falls We want to find out how far it falls in the next 1. The final velocity at the end of the first 1. EXECUTE: a First find the initial speed at the beginning of the first 1. At the end of this 1. This is v0y for the next 1. EVALUATE: The distance the brick falls during the second 1. In this motion the velocity is increasing. In the next EVALUATE: The acceleration is not constant for the entire motion, but it does consist of constant acceleration segments, and we can use constant acceleration equations for each segment. IDENTIFY: When the graph of vx versus t is a straight line the acceleration is constant, so this motion consists of two constant acceleration segments and the constant acceleration equations can be used for each segment. equals the magnitude of the displacement. The acceleration a x is the slope of the vx versus t graph.
The total distance traveled is The graph of a x versus t is given in Figure 2. A collision means the front of the passenger train is at the same location as the caboose of the freight train at some common time. SET UP: Let P be the passenger train and F be the freight train. The collision occurs 0. The equations that specify a collision have a physical solution real, positive t , so a collision does occur. The passenger train moves m before the collision. The freight train moves c The graphs of xF and xP versus t are sketched in Figure 2. IDENTIFY and SET UP: Apply constant acceleration kinematics equations. Find the velocity at the start of the second 5. EXECUTE: For the first 5. This is the initial speed for the second 5. For the second 5. Use this a x and consider the first 5. EVALUATE: The ball is speeding up so it travels farther in the second 5. SET UP: Each car has moved m when they hit.
EVALUATE: The grasshopper ends up m from where it started, so the magnitude of his final displacement is m. This is less than the total distance he travels since he spends part of the time moving in the opposite direction. IDENTIFY: Apply constant acceleration equations to each object. Take the origin of coordinates to be at the initial position of the truck, as shown in Figure 2. Let T be the time it takes the car to catch the truck. EVALUATE: In part c we found that the auto was traveling faster than the truck when they came abreast.
The graph in part d agrees with this: at the intersection of the two curves the slope of the x-t curve for the auto is greater than that of the truck. The auto must have an average velocity greater than that of the truck since it must travel farther in the same time interval. IDENTIFY: The bus has a constant velocity but you have a constant acceleration, starting from rest. SET UP: When you catch the bus, you and the bus have been traveling for the same time, but you have traveled an extra EXECUTE: Call d the distance the bus travels after you start running and t the time until you catch the bus. The distance you must run is This might be possible for a college runner for a brief time, but it would be highly demanding! EVALUATE: Note that when you catch the bus, you are moving much faster than it is. IDENTIFY: Apply constant acceleration equations to each vehicle. SET UP: a It is very convenient to work in coordinates attached to the truck.
Note that these coordinates move at constant velocity relative to the earth. Chapter 2 Motion Along a Straight Line b Need how far the car travels relative to the earth, so go now to coordinates fixed to the earth. Take the origin to be at the initial position of the car. The car travels a IDENTIFY and SET UP: Integrate a x t to find vx t and then integrate vx t to find x t. IDENTIFY: Find the distance the professor walks during the time t it takes the egg to fall to the height of his head. Release the egg when your professor is 3. EVALUATE: Just before the egg lands its speed is 9.
It is traveling much faster than the professor. IDENTIFY: Use the constant acceleration equations to establish a relationship between maximum height and acceleration due to gravity and between time in the air and acceleration due to gravity. IDENTIFY: Calculate the time it takes her to run to the table and return. This is the time in the air for the thrown ball. The thrown ball is in free-fall after it is thrown. Assume air resistance can be neglected. original position. EVALUATE: It takes the ball the same amount of time to reach its maximum height as to return from its maximum height, so when she is at the table the ball is at its maximum height. Note that this large maximum height requires that the act either be done outdoors, or in a building with a very high ceiling.
a IDENTIFY: Consider the motion from when he applies the acceleration to when the shot leaves his hand. SET UP: Take positive y to be upward. c IDENTIFY: Consider the motion of the shot from the point where he releases it to when it returns to the height of his head. EXECUTE: 1. Use the quadratic formula to solve for t:. Since t must be positive, 9. It also takes 0. His head is a little lower than 2. IDENTIFY: The flowerpot is in free-fall. Apply the constant acceleration equations. Use the motion past the window to find the speed of the flowerpot as it reaches the top of the window.
Then consider the motion from the windowsill to the top of the window. The constant-acceleration kinematics formulas all apply. This is the velocity of the 0. Our result says that from the windowsill the pot falls 0. IDENTIFY: Two stones are thrown up with different speeds. a Knowing how soon the faster one returns to the ground, how long it will take the slow one to return? b Knowing how high the slower stone went, how high did the faster stone go? SET UP: Use subscripts f and s to refer to the faster and slower stones, respectively. IDENTIFY: The motion of the rocket can be broken into 3 stages, each of which has constant acceleration, so in each stage we can use the standard kinematics formulas for constant acceleration.
But the acceleration is not the same throughout all 3 stages. During the next before it hits the ground. EVALUATE: We cannot solve this problem in a single step because the acceleration, while constant in each stage, is not constant over the entire motion. The standard kinematics equations apply to each stage but not to the motion as a whole. IDENTIFY: The rocket accelerates uniformly upward at After the engines are off, it moves upward but accelerates downward at 9. Therefore EVALUATE: It we put in 6. IDENTIFY: The teacher is in free-fall and falls with constant acceleration 9. The sound from her shout travels at constant speed.
The sound travels from the top of the cliff, reflects from the ground and then travels upward to her present location. If the height of the cliff is h and she falls a distance y in 3. IDENTIFY: The helicopter has two segments of motion with constant acceleration: upward acceleration for Powers has three segments of motion with constant acceleration: upward acceleration for The quadratic formula 1 After 7. After the next 6. Powers is m above the ground when the helicopter crashes. EVALUATE: When Powers steps out of the helicopter he retains the initial velocity he had in the helicopter but his acceleration changes abruptly from 5. Without the jet pack he would have crashed into the ground at the same time as the helicopter. The jet pack slows his descent so he is above the ground when the helicopter crashes. IDENTIFY: Apply constant acceleration equations to the motion of the rock.
Sound travels at constant speed. SET UP: Let tf be the time for the rock to fall to the ground and let ts be the time it takes the sound to 1 2. travel from the impact point back to you. Both the rock and sound travel a distance h that is equal to the height of the cliff. So you would have overestimated the height of the cliff. It actually takes the rock less time than 8. Quantum Mechanics I: Wave Functions Quantum Mechanics II: Atomic Structure Molecules and Condensed Matter Nuclear Physics Particle Physics and Cosmology. Save my name, email, and website in this browser for the next time I comment. Leave this field empty. com is dedicated to providing trusted educational content for students and anyone who wish to study or learn something new. It is a comprehensive directory of online programs, and MOOC Programs. Terms of Use. Privacy policy. university physics 14th edition instructor solutions manual pdf.
About university physics 14th edition instructor solutions manual pdf Question: Where to find University Physics 14th Edition Instructor Solutions Manual PDF books? Table of Contents of University Physics 14th Edition Instructor Solutions Manual Pdf MECHANICS 1. About the author. The Editorial Team at Infolearners. com is dedicated to providing the best information on learning. From attaining a certificate in marketing to earning an MBA, we have all you need. If you feel lost, reach out to an admission officer. Leave a Comment Cancel reply Comment Name Email Save my name, email, and website in this browser for the next time I comment. About us InfoLearners. Recommended Posts. Since its first edition, University Physics has been revered for its emphasis on fundamental principles and how to apply them. This text is known for its clear and thorough narrative, as well as its uniquely broad, deep, and thoughtful sets of worked examples that provide students with key tools for developing both conceptual understanding and problem-solving skills.
A focus on visual learning, new problem types, and pedagogy informed by MasteringPhysics metadata headline the improvements designed to create the best learning resource for physics students. Chapter 4 and Chapter 22 are available for download as sample chapters in PDF format. Save my name, email, and website in this browser for the next time I comment. Leave this field empty. com is dedicated to providing trusted educational content for students and anyone who wish to study or learn something new. It is a comprehensive directory of online programs, and MOOC Programs. Terms of Use.
It does not measure velocity because it does not measure direction. If the photographs are taken at equal spaced time intervals, then the displacement in successive intervals is increasing and this means the speed is increasing. Therefore, graphs a and e can be ruled out. Graph b shows decreasing acceleration so would correspond to the speed approaching a constant value, which is not what the photographs show. Graph c shows motion in the negative x-direction, which is not the case. This leaves graph d. This graph shows velocity in the positive x-direction and increasing speed.
This is consistent with the photographs. If the object is initially moving and the acceleration direction is opposite to the velocity direction, then the object slows down, stops for an instant and then starts to move in the opposite direction with increasing speed. An example is an object thrown straight up into the air. Gravity gives the object a constant downward acceleration. The object travels upward, stops at its maximum height and then moves downward. The answer to the second question is no. After the first reversal of the direction of travel the velocity and acceleration are then in the same direction.
The object continues moving in the second direction with increasing speed. For an object to be slowing down, all that is required is that the acceleration be nonzero and for the velocity and acceleration to be in opposite directions. The magnitude of the acceleration determines the rate at which the speed is changing. b Yes. For an object to be speeding up, all that is required is that the acceleration be nonzero and for the velocity and acceleration to be in the same direction. But for any nonzero acceleration the speed is increasing when the velocity and acceleration are in the same direction. Average speed is the distance traveled divided by the time interval. Displacement equals the distance traveled when the motion is in the same direction for the entire time interval, and therefore this is when average velocity equals average speed. One average velocity vector is the negative of the other.
The second car could also be observed to be alongside a pedestrian standing at the curb, but that does not mean the pedestrian was speeding. Average velocity is displacement divided by the time interval. If the displacement is zero, then the average velocity must be zero. The answer to the second question is yes. Zero displacement means the object has returned to its starting point, but its speed at that point need not be zero. See Fig. Figure DQ2. An example is a car traveling at constant speed in a straight line. Average acceleration refers to an interval of time and if the velocity is zero throughout that interval, the average acceleration for that time interval is zero.
But yes, you can have zero velocity and nonzero acceleration at one instant of time. For example, in Fig. At its maximum height its velocity is zero but its acceleration is g downward. When the velocity and acceleration are in opposite directions the object is slowing down. The average velocity is the displacement divided by the time interval. The average velocity is defined to be the displacement divided by the time interval. If the acceleration is not constant, objects can have the same initial and final velocities but different displacements and therefore different average velocities. While being thrown, the ball accelerates from rest to velocity v0 y while traveling a distance less than your height. After it leaves your hand, it slows from v0 y to zero at the maximum height, while traveling a distance much greater than your height. Therefore, their speed is continually increasing and the distance one drop travels in each successive 1. A given drop has fallen for 1. The distance between these two drops then is.
The separation Δy increases as t increases. Repeat for successive small time intervals. When it returns to the height from which it was thrown, at the top of the building, it is moving downward with speed v0. The rest of its motion is the same as for the second ball. a Since the last part of the motion of the first ball starts with it moving downward with speed v0 from the top of the building, the two balls have the same speed just before they reach the ground. b The second ball reaches the ground first, since the first ball has to move up and then down before repeating the motion of the second ball. c Displacement is final position minus initial position. Both balls start at the top of the building and end up at the ground. So they have the same displacement. d The first ball has traveled a greater distance. The first The second The average Δx This is less than 4. But its velocity is continually changing, at a constant rate. Acceleration measures the rate of change of velocity.
Also note the comments in part d of the solution to Example 2. To increase y by a factor of 3, increase t by a factor of. The average velocity is zero. EVALUATE: The average velocity for the trip from the nest to the release point is positive. IDENTIFY: Target variable is the time Δt it takes to make the trip in heavy traffic. Use Eq. Now use vav-x for heavy traffic to calculate Δt; Δx is the same as before: Δx This result perhaps surprising occurs because the time interval is inversely proportional to the average speed, not directly proportional to it.
Use the average speed for each segment to find the time Δt traveled in that segment. The average speed is the distance traveled divided by the time. SET UP: The post is 80 m west of the pillar. The average velocity is directed westward. Δt All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher. EVALUATE: The displacement is much less than the distance traveled, and the magnitude of the average velocity is much less than the average speed. The average speed for the entire trip has a value that lies between the average speed for the two segments. IDENTIFY: Given two displacements, we want the average velocity and the average speed. So Δx Use x t to find x for each t. c IDENTIFY: Find the value of t when vx t from part b is zero. Education, All rightsInc. All rights Thisreserved. material This is protected materialunder is protected all copyright under laws all copyright as they currently laws as they exist.
currently exist. No portion of No thisportion material ofmay this be material reproduced, may beinreproduced, any form orinbyany anyform means, or by without any means, permission without in permission writing from inthe writing publisher. from the publisher. IDENTIFY: We know the position x t of the bird as a function of time and want to find its instantaneous velocity at a particular time. dt EVALUATE: The acceleration is not constant in this case. We can find the displacement Δt for each Δt constant velocity time interval. For the first 3. The distance traveled is Δx 7. The average speed is also 2. The ball travels 4. When the motion changes direction during the time interval, those quantities are different. IDENTIFY and SET UP: The instantaneous velocity is the slope of the tangent to the x versus t graph. EXECUTE: a The velocity is zero where the graph is horizontal; point IV.
b The velocity is constant and positive where the graph is a straight line with positive slope; point I.
University Physics 13th Edition Solution Manual,More from Sullivan0502
If you are a student who needs the best University Physics 14th Edition Instructor Solutions Manual Pdf book for physics, you are going to find this website page very helpful. I am sure Download Solution Manual for University Physics With Modern Physics 14th Edition Young Freedman Free in pdf format. Account Login Register Search Search Partner Physics 14th Edition PDF free download book is a comprehensive compilation of Physics concepts that helps you understand various concepts related to Physics easily and effectively. university physics with modern physics 14th edition young freedman Solution manual - Free download as PDF File .pdf), Text File .txt) or read online for free. link download Solution University Physics 13th Edition Solution Manual (PDF) University Physics 13th Edition Solution Manual | Felipe Arias Recio - blogger.com blogger.com no longer supports Internet Explorer blogger.com - Free download as PDF File .pdf), Text File .txt) or read online for free. Scribd is the world's ... read more
com is dedicated to providing the best information on learning. For car A, a x is positive and vx increases with t. The equations that specify a collision have a physical solution real, positive t , so a collision does occur. Mechanical Waves Take the origin of coordinates to be at the initial position of the truck, as shown in Figure 2.
During the next The auto must have an average velocity greater than that of the truck since it must travel farther in the same time interval. SET UP:. c The results can be verified by noting that the x lines for the student and the bus intersect at two points, as shown in Figure 2. EVALUATE: Notice that we were not given the initial speed, but we could find it: 2. Therefore the blood is always flowing forward, but it is increasing in speed during the university physics 14th edition solutions manual pdf free download 0.
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