FUNDAMENTALS OF PHYSICS 10TH EDITION SOLUTION MANUAL. 8 Pages Instructor's Solution Manual for Fundamentals of Physics, 6/E by Halliday, Resnick, and Wa. Strength Hibbeler Solution manual, 8th edition-GearTeam. pdf. The Student Solutions Manual for the 10th edition is written using an innovative approach called TEAL which stands for Think, Express, Analyze, and Learn. Halliday Resnick Walker - Fundamentals of Physics 10th solutions manual. Kong Zahr by clicking the 'Download' button above. Download pdf. ×Close.
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We start with Eq.
We note that the distance x consists of the original gap between them, D, as well as the forward distance traveled during this time by the locomotivev. Therefore, vt v x D v t D v. A graph is shown here for the case where a collision is just avoided x along the vertical axis is in meters and t along the horizontal axis is in seconds.
The top straight line shows the motion of the locomotive and the bottom curve shows the motion of the passenger train. The other case where the collision is not quite avoided would be similar except that the slope of the bottom curve would be greater than that of the top line at the point where they meet.
We are allowed to use Table with y replacing x because this is constant acceleration motion. The ground level is taken to correspond to the origin of the y axis. THINK As the ball travels vertically upward, its motion is under the influence of gravitational acceleration. The kinematics is one-dimensional. Therefore, we can relate its initial speed v0 to y via the equation0 v2 v2 2gy.
The acceleration graph is a horizontal line at —9. At t T 0 , the ball returns to its original position y 0.
This is constant acceleration motion, and we may use Table with y replacing x. We estimate the mass of a raindrop to be about a gram or less, so that its mass and speed from part a would be less than that of a typical bullet, which is good news.
But the fact that one is dealing with many raindrops leads us to suspect that this scenario poses an unhealthy situation. If we factor in air resistance, the final speed is smaller, of course, and we return to the relatively healthy situation with which we are familiar.
This is constant acceleration motion, which justifies the use of Table with y replacing x: LEARN As the wrench falls, with a g 0, its speed increases but its velocity becomes more negative, as indicated by the second graph above.
This is constant acceleration motion, which justifies the use of Table with y replacing x. THINK In this problem a package is dropped from a hot-air balloon which is ascending vertically upward. We analyze the motion of the package under the influence of gravity. This allows us to use Table with y replacing x: The time it takes for the package to hit the ground can be found by solving Eq.
ANALYZE a We solve 0 y y v t 1 gt2 for time using the quadratic0 0 2 formula choosing the positive root to yield a positive value for t: Its final speed is Now we use Eq. We look for conditions such that the key will fall into the boat. Therefore, the speed of the boat is v 12 m 4.
This agrees with our intuition that the lower the height from which the key is dropped, the greater the speed of the boat in order to catch it. We are allowed to use Eq. We use primed variables except t with the first stone, which has zero initial velocity, and unprimed variables with the second stone with initial downward velocity —v0, so that v0 is being used for the initial speed.
SI units are used throughout. THINK The free-falling moist-clay ball strikes the ground with a non-zero speed, and it undergoes deceleration before coming to rest.
EXPRESS During contact with the ground its average acceleration is given by v aavg t , where v is the change in its velocity during contact with the ground and t Consequently, the average acceleration during contact with the ground is a v 0 LEARN Since t is very small, it is not surprising to have a very large acceleration to stop the motion of the ball.
In later chapters, we shall see that the acceleration is directly related to the magnitude and direction of the force exerted by the ground on the ball during the course of collision.
It is then straightforward to solve: To find the velocity just after hitting the floor as it ascends without air friction to a height of 2.
Consequently, the average acceleration is aavg v2 v1 t 6. In a later chapter, this will be directly related to the magnitude and direction of the force exerted by the ground on the ball during the collision. We denote the 1. If we had instead let 0.
The ground level is taken to correspond to the origin of the y-axis. Its position at the moment drop 1 strikes the floor is y 1 gt2 1 9. During its fall, the ball passes with velocity v1 the top of the window which is at y1 at time t1, and passes the bottom which is at y2 at time t2.
From this, Eq. Now Eq. This is a consequence of Eq. As the player reaches y1 We use v for its velocity as it passes the top of the window going up.
From 0 to 40 ms, region A has zero area. From 40 ms to ms, region B has the shape of a triangle with area areaB 1 0. Substituting these values into Eq. From 0 to 10 ms, region A has the shape of a triangle with area The problem is solved using Eq. Carrying out similar calculations for the helmeted head, we have the following results: Thus, the difference in the speed is v vunhelmeted vhelmeted 0. This problem can be solved by noting that velocity can be determined by the graphical integration of acceleration versus time.
The speed of the tongue of the salamander is simply equal to the area under the acceleration curve: To solve this problem, we note that velocity is equal to the time derivative of a position function, as well as the time integral of an acceleration function, with the integration constant being the initial velocity.
Similarly, the velocity of particle 2 is v2 v20 a2dt The condition that v1 v2 implies Thus, the velocity at this time is v1 v2 We adopt the convention frequently used in the text: Therefore, y0 the height of the building is equal to We denote the required time as t, assuming the light turns green when the clock reads zero.
By this time, the distances traveled by the two vehicles must be the same.
Therefore, a 2. If the plane with velocity v maintains its present course, and if the terrain continues its upward slope of 4. We denote tr as the reaction time and tb as the braking time. The motion during tr is of the constant-velocity call it v0 type. Then the position of the car is given by 1 2 x v0tr v0tb atb 2 where v0 is the initial velocity and a is the acceleration which we expect to be negative-valued since we are taking the velocity in the positive direction and we know the car is decelerating.
The magnitude of the deceleration is therefore 6. Note that we could have chosen Eq. The deceleration is constant so that Table can be used. Since the primed train has the lower initial speed, it should stop sooner than the other train would were it not for the collision. If the computation of v had failed meaning that a negative number would have been inside the square root then we would have looked at the possibility that there was no collision and examined how far apart they finally were.
A concern that can be brought up is whether the primed train collides before it comes to rest; this can be studied by computing the time it stops Eq. We use subscripts 1 and 2 for the data. Thus, we are asked to find the value of x1.
Although any of several equations could be used, we choose Eq. An integration is required to calculate velocity. In our situation, we have a 5. Since the acceleration is linear in t, we expect the velocity to be quadratic in t, and the displacement to be cubic in t. Thus, fort1 We express this as a multiple of g by setting up a ratio: Let D be the distance up the hill.
We obtain the velocity by integration of the acceleration: Lengths are in meters and times are in seconds. The student is encouraged to look at the discussion in Section to better understand the manipulations here.
This is the correct answer, but one has the right to worry that it might not be; after all, the problem asks for the total distance traveled and x x0 is just the displacement. If the cyclist backtracked, then his total distance would be greater than his displacement. Thus, we might ask, "did he backtrack? Driving at higher speed within the legal limit reduces travel time. The acceleration is constant and we may use the equations in Table Thus, t v0 5.
THINK In this problem we explore the connection between the maximum height an object reaches under the influence of gravity and the total amount of time it stays in air. Note also that for t 2t , the initial speed must be twice the original speed: Thus, the top of its motion is 1. The total time of fall can be computed from Eq. THINK This problem involves analyzing a plot describing the position of an iceboat as function of time.
The boat has a nonzero acceleration due to the wind. However, the challenge here is that v0, v, and a are not explicitly given.
Our strategy to deduce these values is to apply the kinematic equation x x v t 1 at2 to a variety of points on the graph and solve for the0 0 2 unknowns from the simultaneous equations. LEARN By using the results obtained in a , the position and velocity of the iceboat as a function of time can be written as x t 6. One can readily verify that the same answers are obtained for b and c using the above expressions for x t andv t.
Let the depth of the lake be D, and the total time for the ball to descend be T. Now, the time it spends descending in the lake at constant velocity v is t D D. If, however, upward had been chosen as the positive direction, then this answer in b would turn out negative-valued.
Thus, h D gT 5. We use subscript 2 for the elevator reaching the ground and 1 for the halfway point. We are starting the clock when the first1 2 object is dropped. Therefore, using Eq. The initial velocity is zero so that Eq. Thus, we can write the equation as d 1 gt2 for simplicity. This is a result of Eq. The positive root is taken because the problem asks for the speed the magnitude of the velocity. An important perspective related to this is treated later in the book in the context of energy conservation.
The details follow: We assume constant velocity motion and use Eq. Assuming the horizontal velocity of the ball is constant, the horizontal displacement is x vt , where x is the horizontal distance traveled, t is the time, and v is the horizontal velocity. Thus t x Its slope is the average velocity during the first 3s of motion. The problem consists of two constant-acceleration parts: And Eq.
We could find the distance for part 2 from several of the equations, but the one that makes no use of our part a results is Eq.
Therefore, the total distance traveled by the shuffleboard disk is 1. The time required is found from Eq.
First, we convert the velocity change to SI units: From Table , v2 v2 2ax is used to solve for a. This problem consists of two parts: The equations for parts 1 and 2, used above, therefore become x 0 1 0 Turning to the constant acceleration equations in Table , we use x 1 v v t.
Thus, Thus, the speed is v The negative value of y2 signifies that the distance traveled while arresting its motion is downward. The total time elapsed is t 2 h 41 min min and the center point is displaced by x 3. Thus, the average velocity of the center point is v x cm 2. Bookmark it to easily review again before an exam. The best part?
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