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We are pleased to be able to provide you \n \n with a variety of support material to help you in the teaching of your course.Before putting any of the solutions on a \n \n Web site, we ask that you request formal permission from us to do so. In most cases we will grant such permission PROVIDED \n \n THAT THE WEB SITE IS PASSWORD PROTECTED. \n \n \n Our goal is to prevent students from other campuses from being able to access \n \n your solutions. Thus, R is less than the magnitude \n \n (length) of A plus the magnitude of B. \n \n \n \n 3. (a) The triangle in the drawing is a right triangle. Thus, the resultant vector is zero. \n \n \n \n 8. (c) When the two vector components A x and A y are added by the tail-to-head method, the \n \n sum equals the vector A. These units differ from the given units in \n \n both length and time dimensions, so both must be converted. Therefore, we can solve the equation for z and then substitute the \n \n known dimensions. SSM REASONING The shortest distance between the two towns is along the line that \n \n joins them. REASONING The drawing shows a schematic \n \n representation of the hill. REASONING In both parts of the drawing the line of sight, the horizontal dashed line, and \n \n the vertical form a right triangle. The bottom face of the \ncube is a square whose diagonal has a length L. REASONING \n \n a. The drawing shows the person standing on the \n \n earth and looking at the horizon. SSM REASONING \n \n a. Since the two force vectors A and B have directions due west and due north, they are \n \n perpendicular. The direction of the second force \n \n is not specified; it could point either due east or due west, so there are two answers. SSM REASONING For convenience, we can assign due east to be the positive direction \n \n and due west to be the negative direction. Since all the vectors point along the same east-\n \n west line, the vectors can be added just like the usual algebraic addition of positive and \n \n negative scalars. REASONING At the turning point, the distance to the \n \n campground is labeled d in the drawing. Note that d is \n \n the length of the hypotenuse of a right triangle. REASONING The triple jump consists of a double jump in one direction, followed by a \n \n perpendicular single jump, which we can represent with \n \n displacement vectors J and K (see the drawing). In \n \n order to find the magnitude D of the displacement, we first need \n \n to find the magnitudes J and K of the double jump and the single \n \n jump. REASONING Both P and Q and the \n \n vector sums K and M can be drawn with \n \n correct magnitudes and directions by \n \n counting grid squares. To add vectors, \n \n place them tail-to-head and draw the \n \n resultant vector from the tail of the first \n \n vector to the head of the last. SSM REASONING AND SOLUTION The single force needed to produce the same \n \n effect is equal to the resultant of the forces provided by the two ropes. The following \n \n figure shows the force vectors drawn to scale and arranged tail to head. REASONING a. Since the two displacement vectors A and B have directions due south \n \n and due east, they are perpendicular. REASONING AND SOLUTION The following figure is a scale diagram of the forces \n \n drawn tail-to-head. The scale factor is shown in the figure. Likewise, the magnitude of \n \n the y component of the force vector is the product of the magnitude of the force times the \n \n sine of the angle between the vector and the x axis. SSM REASONING AND SOLUTION In order to determine which vector has the \n \n largest x and y components, we calculate the magnitude of the x and y components \n \n explicitly and compare them.REASONING The triangle in the drawing is a right triangle.SSM REASONING The x and y components of r are mutually perpendicular; therefore, \n \n the magnitude of r can be found using the Pythagorean theorem. REASONING Two vectors that are equal must have the same magnitude and direction. \n \n Equivalently, they must have identical x components and identical y components. We will \n \n begin by examining the given information with respect to these criteria, in order to see if \n \n there are obvious reasons why some of the vectors could not be equal. To verify that this is \n \n indeed the case we have two choices. We can either calculate the magnitude and direction \n \n of A (and compare it to the given magnitude and direction of C ) or determine the scalar \n \n components of C (and compare them to the given components of A ). SSM REASONING The force F and its two components form a right triangle. REASONING AND SOLUTION The force F can be first resolved into two components; \n \n the z component Fz and the projection onto the x - y plane, Fp as shown on the left in the \n \n following figure. SSM REASONING The individual displacements of the golf ball, A, B, and C are \n \n shown in the figure. REASONING To apply the component method for vector addition, we must first determine \n \n the x and y components of each vector. The algebraic sum of the three x components gives \n \n the x component of the resultant. The algebraic sum of the three y components gives the \n \n y component of the resultant. Knowing the x and y components of the resultant will allow \n \n us to use the Pythagorean theorem to determine the magnitude of the resultant. The drawing \n \n shows these two components and the resultant \n \n vector. REASONING Using the component method for vector addition, we will find the x \n \n component of the resultant force vector by adding the x components of the individual \n \n vectors. Then we will find the y component of the resultant vector by adding the y \n \n components of the individual vectors. Once the x and y components of the resultant are \n \n known, we will use the Pythagorean theorem to find the magnitude of the resultant and \n \n trigonometry to find its direction. Because \n \n FB and FC are directed symmetrically about the x axis, and have the same magnitude F, \n \n their y components are equal and opposite. Therefore, they cancel out of the vector sum R, \n \n leaving only the x components of FB and FC. REASONING Using the component method, we find the components of the resultant R \n \n that are due east and due north. The magnitude and direction of the resultant R can be \n \n determined from its components, the Pythagorean theorem, and the tangent function. \n \n \n \n SOLUTION The first four rows of the table below give the components of the vectors A, \n \n B, C, and D. REASONING We will use the scalar x and y components of the resultant vector to obtain \n \n its magnitude and direction. To obtain the x component of the resultant we will add together \n \n the x components of each of the vectors. Once we have these \n \n components, we can calculate the magnitude and direction of C, as shown on the GPS \n \n readout. SSM REASONING Since the finish line is coincident with the starting line, the net \n \n displacement of the sailboat is zero. Hence the sum of the components of the displacement \n \n vectors of the individual legs must be zero. REASONING The following table shows the components of the individual displacements \n \n and the components of the resultant. REASONING The drawing shows the vectors A, \n \n B, and C. These \n \n two equations will allow us to find the magnitudes of \n \n B and C. \n \n \n \n \n \n \n \n \n \n \n \n \n \n SOLUTION The x - and y -components of A, B, and C are given in the table below. This means that the sum of the x components of the vectors and the sum \n \n of the y components of the vectors are separately equal to zero. From these two equations \n \n we will be able to determine the magnitudes of vectors B and C. SSM REASONING The ostrich's velocity vector v and \n \n the desired components are shown in the figure at the right. \n \n The components of the velocity in the directions due west and \n \n due north are v\n W\n \n and v\n N\n \n, respectively. REASONING AND SOLUTION The east and north components are, respectively \n \n \n \n a. REASONING According to the component \n \n method for vector addition, the x component of the \n \n resultant vector is the sum of the x component of A \n \n and the x component of B. Similarly, the y \n \n component of the resultant vector is the sum of the \n \n y component of A and the y component of B. The \n \n magnitude R of the resultant can be obtained from \n \n the x and y components of the resultant by using \n \n the Pythagorean theorem. SSM REASONING The performer walks out on the wire a distance d, and the vertical \n \n distance to the net is h. Values for s and h are \n \n given, so we can solve this expression for the distance d. Vectors A and B are the components of the resultant, C. REASONING AND SOLUTION The following figures are scale diagrams of the forces \n \n drawn tail-to-head. REASONING There are two right \n \n triangles in the drawing. Each \n \n contains the common side that is \n \n shown as a dashed line and is \n \n labeled D, which is the distance \n \n between the buildings. The \n \n hypotenuse of each triangle is one \n \n of the lines of sight to the top and \n \n base of the taller building. Since the average acceleration is the change in \n \n velocity divided by the elapsed time, the average acceleration is also zero. \n \n 9. (a) The runners are always moving after the race starts and, therefore, have a non-zero \n \n average speed. The average velocity is the displacement divided by the elapsed time, and the \n \n displacement is zero, since the race starts and finishes at the same place. The average \n \n acceleration is the change in the velocity divided by the elapsed time, and the velocity \n \n changes, since the contestants start at rest and finish while running. \n \n 10. (c) The equations of kinematics can be used only when the acceleration remains constant \n \n and cannot be used when it changes from moment to moment. \n \n 11. (a) Velocity, not speed, appears as one of the variables in the equations of kinematics. \n \n Velocity is a vector. While the rocket is picking up speed in the upward direction, the acceleration is not \n \n just due to gravity, but is due to the combined effect of gravity and the engines. In fact, the \n \n effect of the engines is greater than the effect of gravity. Therefore, this stone picks up speed as \n \n it approaches the nest. In contrast, the acceleration due to gravity points opposite to the \n \n initial velocity of the stone thrown from the ground, so that this stone loses speed as it \n \n approaches the nest. The result is that, on average, the stone thrown from the top of the cliff \n \n travels faster than the stone thrown from the ground and hits the nest first. \n \n 23. 1.13 s \n\n \n 44 KINEMATICS IN ONE DIMENSION \n \n 24. (a) The slope of the line in a position versus time graph gives the velocity of the motion. \n \n The slope for part A is positive. For part B the slope is negative. For part C the slope is \n \n positive. \n \n 25. (b) The slope of the line in a position versus time graph gives the velocity of the motion. \n \n Section A has the smallest slope and section B the largest slope. \n \n 26. (c) The slope of the line in a position versus time graph gives the velocity of the motion. \n \n Here the slope is positive at all times, but it decreases as time increases from left to right in \n \n the graph. If the final position is greater than the initial position, the displacement is \n \n positive. Distance is a \n \n scalar, and displacement is a vector. Distance and the magnitude of the displacement, \n \n however, are both measured in units of length. \n \n \n \n SOLUTION \na. REASONING AND SOLUTION Let west be the positive direction. SSM REASONING AND SOLUTION \n \n a. The total displacement traveled by the bicyclist for the entire trip is equal to the sum of \n \n the displacements traveled during each part of the trip. The displacement in this expression is the \n \n total displacement, which is the sum of the displacements for each part of the trip. \n \n Displacement is a vector quantity, and we must be careful to account for the fact that the \n \n displacement in the first part of the trip is north, while the displacement in the second part is \n \n south. \n \n \n \n SOLUTION According to Equation 2.2, the displacement for each part of the trip is the \n \n average velocity for that part times the corresponding elapsed time. Note that the \n \n minus sign indicates a direction due south. The change in velocity \nis equal to the final velocity minus the initial velocity. Therefore, the change in velocity, and \n \n hence the acceleration, is positive if the final velocity is greater than the initial velocity. The \n \n acceleration is negative if the final velocity is less than the initial velocity. REASONING Although the planet follows a curved, two-dimensional path through space, \n \n this causes no difficulty here because the initial and final velocities for this period are in \n \n opposite directions. REASONING Since the velocity and acceleration of the motorcycle point in the same \n \n direction, their numerical values will have the same algebraic sign. For convenience, we \n \n will choose them to be positive. REASONING When the velocity and acceleration vectors are in the same direction, the \n \n speed of the object increases in time. REASONING The fact that the emu is slowing down tells us that the acceleration and the \n \n velocity have opposite directions. Since the bird is slowing down, its \n \n acceleration must point in the opposite direction, or due south. \n \n \n \n b. We assume that due north is the positive direction. REASONING AND SOLUTION Both motorcycles have the same velocity v at the end of \n \n the four second interval. REASONING The average acceleration is defined by Equation 2.4 as the change in \n \n velocity divided by the elapsed time. REASONING We know the initial and final velocities of the blood, as well as its \n \n displacement. The actual \n \n time required for Secretariat to run the final mile can be determined from Equation 2.8, \n\n \n 58 KINEMATICS IN ONE DIMENSION \n \n since the initial velocity, the acceleration, and the displacement are given. In other words, the cart is decelerating, and its acceleration must point \n \n opposite to the velocity, or to the left. Thus, the acceleration is negative. REASONING At time t both rockets return to their starting points and have a displacement \n \n of zero. REASONING The stopping distance is the sum of two parts. According to Equation 2.2, this \n \n distance is the magnitude of the displacement and is the magnitude of the velocity times the \n \n time. Second, there is the distance the car travels while it decelerates as the brakes are \n \n applied. Since it is decelerating, its \n \n acceleration points opposite to its velocity. REASONING The drawing shows the two knights, initially separated by the displacement \n \n d, traveling toward each other. REASONING The players collide when they have the same x coordinate relative to a \n \n common origin. For convenience, we will place the origin at the starting point of the first \n \n player. This is also the travel time for the second car to reach the next \n \n exit. The acceleration for the second car can be determined from Equation 2.8, since the \n \n initial velocity, the displacement, and the time are known. REASONING Let the total distance between the first and third sign be equal to 2 d. SSM REASONING AND SOLUTION When air resistance is neglected, free fall \n \n conditions are applicable. REASONING AND SOLUTION The figure at the right \n \n shows the paths taken by the pellets fired from gun A and \n \n gun B. The two paths differ by the extra distance covered \n \n by the pellet from gun A as it rises and falls back to the \n \n edge of the cliff. When it falls back to the edge of the \n \n cliff, the pellet from gun A will have the same speed as \n \n the pellet fired from gun B, as Conceptual Example 15 \n \n discusses. REASONING The initial velocity and the elapsed time are given in the problem. Once the initial speed of the ball is known, Equation 2.9 can be used a \nsecond time to determine the height above the launch point when the speed of the ball has \n \n decreased to one half of its initial value. \n \n \n \n SOLUTION When the ball has reached its maximum height, its velocity is zero. SSM REASONING AND SOLUTION Since the balloon is released from rest, its initial \n \n velocity is zero. REASONING The ball is initially in free fall, then collides with the pavement and \n \n rebounds, which puts it into free fall again, until caught by the boy. For each of the intervals, the acceleration is that due to gravity.