Giancoli physics 7th edition solutions pdf free download

Let the origin be the location of the fugitive when he starts to a 1. Thus the fugitive cannot catch the car during his 4. Now the equation of motion of the fugitive changes. After the 4. For the runners to cross the finish line side-by-side, they must both reach the finish line in the same amount of time from their current positions.

For Sally:. The first time is the time she first crosses the finish line, so that is the time to be used for the problem. For Mary:. Define the origin to be the location where the speeder passes the police car. Start a timer at the instant that the speeder passes the police car and find another time that both cars have the same displacement from the origin. The first part is the distance traveled at the initially constant speed during the 1 second of reaction time.

So this second part of the police car displacement, using Eq. The reason it is not exactly zero is rounding of previous values. The answer of The displacement is found from Eq. The displacement upon catching the ball is 0, assuming it was caught at the same height from which it was thrown. The starting speed can be found from Eq. The height can be calculated from Eq. The time of flight can be found from Eq.

This is an estimate primarily because the effects of the air have been ignored. So ignoring the effects of air makes this an estimate. Finally, we assume that the ball was caught at the same height as which it was hit. That was not stated in the problem either, so that is an estimate. Consider just the downward motion of the kangaroo. The total time is then twice the falling time. The speed of the object will be given by Eq. This is the equation of a straight line passing through the origin with a slope of 9.

The distance fallen will be given by Eq. This is the equation of a parabola, with its vertex at the origin, opening upward. Then we can calculate the position as a function of time from Eq. The distance traveled during each second can be found by subtracting two adjacent position values from the above list. Let n be a positive integer, starting with 0. Find the velocity of the rocket when it runs out of fuel from Eq. Note that the value has 2 significant figures.

The time to reach the m location can be found from Eq. The altitude reached from the out-of-fuel point can be found from Eq. Find the velocity upon reaching the Earth from Eq. The time to reach the ground can be found from Eq. Using Eq. At the top of its path, the velocity will be 0, so we can use the initial velocity as found above, along with Eq.

Call the location of the top of the window yw , and the time for the stone to fall from release to the top of the window is tw. Since the stone is dropped from rest, using Eq.

The location of the bottom of the 2. Use this time in the first equation to find the desired distance. Insert this expression for t1 into the equation for H from the stone, and solve for H.

The indication of a constant velocity on a velocity vs. The indication of a constant acceleration on a velocity vs. Slightly different answers may be obtained since the data come from reading the graph. The object moves in both directions. The height at the top of the path can be found from Eq. From this we see that the displacement is inversely proportional to the acceleration, so if the acceleration is reduced by a factor of 6 by going to the Moon, and the initial velocity is unchanged, then the displacement increases by a factor of 6.

Find the displacement from Eq. Find the speed upon reaching the net from Eq. The initial velocity is. For the acceleration to be smaller, in the above equation we see that the displacement would have to be larger. This means that the net should be loosened. The fish can take evasive action if he sees the pelican at a time of 1. Find the location of the pelican at that time from Eq.

Find the average acceleration from Eq. The time to travel the m is found using the distance and the speed limit. No , you cannot make it to the third light without stopping, since it takes you longer than The distance traveled during that time is found from Eq.

Since 5. The car travels another 7. Thus the total distance is No , the car cannot make it through all three lights without. The car has to travel another Note that Eq. Find the final velocity from Eq. Find the initial velocity of the ball from Eq. The range of acceptable velocities for the uphill lie is 4.

The range of acceptable velocities for the downhill lie is 5. The total distance traveled will be the distance up plus the distance down. The distance down will be 75 m more than the distance up.

To find the distance up, use the fact that the speed at the top of the path will be 0. Then using Eq. Thus the distance up is This problem can be analyzed as a series of three one-dimensional motions: the acceleration phase, the constant-speed phase, and the deceleration phase. Find the elapsed time for the acceleration phase from Eq. Find the displacement during the acceleration phase from Eq.

Find the elapsed time for the deceleration phase from Eq. Find the distance traveled during the deceleration phase from Eq. A train making the entire trip would thus have a total of 5 interstation segments and 4 stops of 22 s each at the intermediate stations. With 5 interstation segments of A train making the entire trip would thus have a total of 3 interstation segments and 2 stops of 22 s each at the intermediate stations.

With 3 interstation segments of The 2. Case II: crossing the intersection. The critical condition is that the total distance covered by the passing car and the approaching car must be less than m so that they do not collide. The passing car has a total displacement composed of several individual parts. These are i the 10 m of clear room at the rear of the truck, ii the m length of the truck, iii the 10 m of clear room at the front of the truck, and iv the distance the truck travels.

To express the motion of the car, we choose the origin to be at the location of the passing car when the decision to pass is made. Calculate the displacements of the two cars during this time. Thus the two cars together have covered a total distance of The car should not pass.

Find the time of fall from Eq. Choose downward to be the positive direction and the origin to be at the top of the building. The difference in the estimates is 6. If we assume the height of the building is the average of the 6. The intent of the method was probably to use the change in air pressure between the ground level and the top of the building to find the height of the building.

The very small difference in time measurements, which could be due to human reaction time, makes a 6. This could be as much as 2 floors in error. The two bicycles will have the same velocity at any time when the instantaneous slopes of their x vs. That occurs near the time t1 as marked on the graph. Bicycle A has the larger acceleration, because its graph is concave upward, indicating a positive acceleration.

Bicycle B has no acceleration because its graph has a constant slope. The graph with the steepest slope is the faster bicycle, so it is the one that is passing at that instant. So at the first crossing, bicycle B is passing bicycle A. At the second crossing, bicycle A is passing bicycle B. Bicycle B has the highest instantaneous velocity at all times until the time t1 , where both graphs have the same slope. For all times after t1 , bicycle A has the highest instantaneous velocity. The largest instantaneous velocity is for bicycle A at the latest time shown on the graph.

The bicycles appear to have the same average velocity. If the starting point of the graph for a particular bicycle is connected to the ending point with a straight line, the slope of that line is the average velocity.

To find the average speed for the entire race, we must take the total distance divided by the total time. If one lap is a distance of L, then the total distance will be 10L. The time elapsed at a given constant 9L , and the time for Choose upward to be the positive direction.

Since the first child can bounce up to one-and-a-half times higher than the second child, the first child can bounce up to a height of 1. To find the time that the first child was in the air, use Eq. The time of 0 s corresponds to the time the child started the jump, so the correct answer is 1. Choose downward to be the positive direction and the origin to be at the location of the plane.

As shown in Example 2—15, the speed with which the ball was thrown upward is the same as its speed on returning to the ground. Choose upward to be the positive direction and the origin to be at the level where the ball was thrown.

Find the initial velocity from Eq. The two conditions are that the motion needs to be near the surface of the Earth and that there is no air resistance. An example where the second condition is not even a reasonable approximation is that of parachuting. The air resistance caused by the parachute results in the acceleration not being constant, with values much different than 9. The sounds will not occur at equal time intervals because the longer any particular bolt falls, the higher its speed.

With equal distances between bolts, each successive bolt, having fallen a longer time when its predecessor reaches the plate, will have a higher average velocity and thus travel the interbolt distance in shorter periods of time.

Thus the sounds will occur with smaller and smaller intervals between sounds. To hear the sounds at equal intervals, the bolts would have to be tied at distances corresponding to equal time intervals. The first bolt call it bolt 0 is touching the plate. If the distance from the bolt initially on the pan to the next bolt is d1 , then the distance from that bolt to the next one is 3d1 , the distance to the next bolt is 5d1 , and so on.

The accompanying table shows these relationships in a simpler format. The location of the police car as a function of time is given by Eq. The time to overtake the speeder occurs when the speeder has gone a distance of m.

During the interval from A to B, it is moving in the negative direction , because its displacement is negative. Chapter 13 Temperature and Kinetic Theory 69 solutions. Chapter 14 Heat 45 solutions. Chapter 15 The Laws of Thermodynamics 55 solutions. Chapter 16 Electric Charge and Electric Field 43 solutions. Chapter 17 Electric Potential 67 solutions.

Chapter 18 Electric Currents 60 solutions. Chapter 19 DC Circuits 66 solutions. Chapter 20 Magnetism 63 solutions. Chapter 21 Electromagnetic Induction and Faraday's Law 73 solutions. Chapter 22 Electromagnetic Waves 43 solutions.

Chapter 23 Light: Geometric Optics 68 solutions. Chapter 24 The Wave Nature of Light 72 solutions. Chapter 25 Optical Instruments 63 solutions. Chapter 26 Special Theory of Relativity 53 solutions. Chapter 28 Quantum Mechanics of Atoms 44 solutions.