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For convenience, we will choose them to be positive. The velocity, acceleration, and the time are related by Equation 2. Solving Equation 2. When the velocity and acceleration vectors are in opposite directions, the speed of the object decreases in time. The final velocities and speeds for the four moving objects are: a. The final speed is 6. The initial velocity of the emu is directed due north. Since the bird is slowing down, its acceleration must point in the opposite direction, or due south.

We assume that due north is the positive direction. With the data given for the first part of the motion, Equation 2. The magnitude of the acceleration can be found from Equation 2. Assuming that the acceleration is constant, the displacement covered by the car can be found from Equation 2. We can find the elapsed time from this relation because the acceleration and the change in velocity are given. Therefore, Equation 2. With x representing the displacement traveled along the slope, Equation 2.

The actual time required for Secretariat to run the final mile can be determined from Equation 2. It is the difference between these two results for the time that we seek. The acceleration allowed Secretariat to run the last mile in a time that was faster by In other words, the cart is decelerating, and its acceleration must point opposite to the velocity, or to the left.

Thus, the acceleration is negative. This occurs, because each rocket is decelerating during the first half of its journey. Chapter 2 Problems 59 However, rocket A has a smaller initial velocity than rocket B. Therefore, in order for rocket B to decelerate and return to its point of origin in the same time as rocket A, rocket B must have a deceleration with a greater magnitude than that for rocket A. Since we know that the displacement of each rocket is zero at time t, since both initial velocities are given, and since we seek information about the acceleration, we begin our solution with Equation 2.

First, there is the distance the car travels at According to Equation 2. Second, there is the distance the car travels while it decelerates as the brakes are applied. This distance is given by Equation 2. Since it is decelerating, its acceleration points opposite to its velocity. This velocity is found from Equation 2. Since both cars begin and end the interval side-by-side, they both undergo the same displacement x during this interval.

The displacement of each car is given by Equation 2. For convenience, we will place the origin at the starting point of the first player. This is also the travel time for the second car to reach the next exit. The acceleration for the second car can be determined from Equation 2. This equation applies, because the acceleration is constant. None of these four equations contains all three known quantities and the desired displacement x, and each of them contains the initial velocity v0.

Since the initial velocity is neither known nor requested, we can combine two kinematic equations to eliminate it, leaving an equation in which x is the only unknown quantity. Once v is known, Equation 2. Since the train starts from rest with a constant acceleration, the displacement of the train in a time t is given by Equation 2. The final speed can be found from Equation 2.

In a , we will consider the interval beginning at launch and ending 2. In b , we will consider the interval beginning at launch and ending 5. Therefore, we may use Equation 2. The two paths differ by the extra distance covered by the pellet from gun A as it rises and falls back to the edge of the cliff. When it falls back to the edge of the cliff, the pellet from gun A will have the same speed as the pellet fired from gun B, as Conceptual Example 15 discusses.

Therefore, the flight time of pellet A will be greater than that of B by the amount of time that it takes for pellet A to cover the extra distance. Using this information, we can employ Equation 2. The initial position of the compass is 3. Equation 2. The second solution, being a negative time, is discarded.

Once the initial speed of the ball is known, Equation 2. If we take upward as the positive direction, we have from Equation 2. The diver's velocity is zero at his highest point. However, we will need to know 1 2 the initial speed v0, which can be determined via Equation 2. The time required to fall through a vertical displacement y can be found from Equation 2.

This initial speed can be found by applying Equation 2. Thus, according to Equation 2. To determine her velocity at the beginning of the 1. The acceleration is defined by Equation 2. The acceleration is negative because it points downward, and this direction is the negative direction. The displacement of the diver during the last 1. For each of the intervals, the acceleration is that due to gravity.

However, the symmetry of free-fall motion is such that it takes the ball as much time to rise from the ground to a maximum height h2 as it would take for a ball dropped from h2 to fall to the pavement, so we can again use Equation 2. Remember, we are using symmetry to treat the second interval as if the ball were dropped from rest at a height of 5. The speed is the distance divided by the time, according to Equation 2. The distance is 7. The time is the time it takes for the stone to fall, which can be obtained from Equation 2.

The stone that is thrown downward gains speed on the way down. The stones cross paths below the point that corresponds to half the height of the cliff. To see why, consider where they would cross paths if they each maintained their initial speed as they moved. Then, they would cross paths exactly at the halfway point. However, the stone traveling upward begins immediately to lose speed, while the stone traveling downward immediately gains speed.

Thus, the upward moving stone travels more slowly than the downward moving stone. Consequently, the stone thrown downward has traveled farther when it reaches the crossing point than the stone thrown upward.

The initial velocity v0 is known for both stones, as is the acceleration a due to gravity. In addition, we know that at the crossing point the stones are at the same place at the same time t. Furthermore, the position of each stone is specified by its displacement y from its starting point. In using this equation, we will assume upward to be the positive direction.

Note that yup is the displacement of the upward H moving stone above the base of the cliff, ydown is the displacement of the downward moving stone below the top yup of the cliff, and H is the displacement of the cliff-top above the base of the cliff, as the drawing shows. We can use Equation 2. In the previous calculation the acceleration due to gravity was eliminated algebraically. Thus, a value other than 9. Using this fact, we will be able to determine the time it takes for each arrow to reach its maximum height.

Knowing this time for the second arrow will allow us to determine its initial speed at launch. Therefore, since both arrows reach their maximum height at the same time, the second arrow reaches its maximum height 2. The initial speed of the second arrow can then be found from Equation 2. The velocity of the block at the instant that the man looks up can be determined from Equation 2.

Once the velocity is known at that instant, Equation 2. Thus, t is the solution to the quadratic equation 4. We will take upward as the positive direction. Its velocity at the end of the burn, v1, is also the initial velocity for the second portion of its flight: engine burn-out to maximum altitude.

We solve Equation 2. The time t3. The tile falls from rest, so its initial velocity v0 is zero. The tile falls through a displacement y in going from the roof top to the top of the window. The velocity vwindow at the top of the window is not given, but it can be obtained from the time of 0. The algebraic sign of the average velocity, therefore, corresponds to the sign of the slope. The slope, and hence the average velocity, is positive for segments A and C, negative for segment B, and zero for segment D.

In the given position-versus-time graph, we find the slopes of the four straight-line segments to be 1. Each of the three segments of the graph is a straight line, so the bus has a different constant velocity for each part of the trip: vA, vB, and vC. During each ten-second segment, the runner has a constant velocity and runs Chapter 2 Problems 83 half the remaining distance to the finish line. Since the runner has a constant velocity during each ten-second segment, we can find the velocity during each segment from the slope of the position-time graph for that segment.

The following figure shows the position-time graph for the first forty seconds. Each runs ten meters during the first second and, during each second thereafter, each runner runs ninety percent of the distance he ran in the previous second. While the velocity of each runner changes from second to second, it remains constant during any one second.

SOLUTION The following table shows the distance covered during each second for one of the runners, and the position at the end of each second assuming that he begins at the origin for the first eight seconds. Time t s Distance covered m Position x m Chapter 2 Problems 0. Since the two runners are running toward each other in exactly the same way, they will meet halfway between their respective starting points. Cutnell , tutorials, pdf, ebook, torrent, downloads, rapidshare, filesonic, hotfile, megaupload, fileserve.

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Authors: Alberts B. As the amount of information in biology expands dramatically, it becomes increasingly important for textbooks to distill the vast amount of scientific knowledge into concise principles and enduring concepts. As with previous editions, Molecular McGraw-Hill, A comprehensive text with enhanced calculus coverage incorporating a consistently used 7-step problem solving method.

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