College Physics: A Strategic Approach, 3rd Edition. Randall D. Knight, (Professor Emeritus), California Polytechnic State University-San Luis Obispo. College Physics: A Strategic Approach, 4th Edition. Randall D. Knight, (Professor Emeritus), California Polytechnic State University-San Luis Obispo. Instructor's Solutions Manual (Download Only) for College Physics: A Strategic Approach, 3rd Edition. Randall D. Knight, (Professor Emeritus), California.
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Full file at ronaldweinland.info Edition-Knight-Solutions-Manual MOTION IN ONE DIMENSION 2 Q Reason: . ronaldweinland.info Physics for Engineers and Scientists -- 3rd ed. Vol. ronaldweinland.info Uploaded by. Minh Hằng College Physics: A Strategic Approach 2e Knight Solutions Manual. Book Details Author: Randall D. Knight (Professor Emeritus),Brian Jones,Stuart Field Pages: Binding: Hardcover Brand: Addison-Wesley Educational Publishers Inc ISBN: Download or read College Physics: A Strategic Approach by click link below Download or read.
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Randall D. Download instructor resources. Additional order info. Pearson offers special pricing when you package your text with other student resources. If you're interested in creating a cost-saving package for your students, contact your Pearson rep.
On-line Supplement. We're sorry! We don't recognize your username or password. Please try again. The work is protected by local and international copyright laws and is provided solely for the use of instructors in teaching their courses and assessing student learning. You have successfully signed out and will be required to sign back in should you need to download more resources. The other car will pass the origin at a later time t1 and passes you at time t2.
The slope of the position graph is the velocity, and the slope for the faster car is steeper. The acceleration vector will point south when the car is slowing down while traveling north.
The acceleration vector will always point in the direction opposite the velocity vector in straight line motion if the object is slowing down. Feeling good about this concept requires letting go of the common every day mis usage where velocity and acceleration are sometimes treated like synonyms. Physics definitions of these terms are more precise and when discussing physics we need to use them precisely.
A predator capable of running at a great speed while not being capable of large accelerations could overtake slower prey that were capable of large accelerations, given enough time. However, it may not be as effective as surprising and grabbing prey that are capable of higher acceleration. For example, prey could escape if the safety of a burrow were nearby.
If a predator were capable of larger accelerations than its prey, while being slower in speed than the prey, it would have a greater chance of surprising and grabbing prey, quickly, though prey might outrun it if given enough warning. Consider the horse-man race discussed in the text. We will neglect air resistance, and thus assume that the ball is in free fall. Students would be wise to dwell on this question until it makes complete sense.
While in free fall, the acceleration of the rock is exactly the acceleration due to gravity, which has a magnitude g and is downward. This does not affect the acceleration of gravity, which does not depend on how the rock was thrown. Its acceleration remains the acceleration of gravity. Its velocity has increased due to gravity, but acceleration due to gravity is independent of velocity. No matter what the velocity of an object is, the acceleration due to gravity always has magnitude g and is always straight downward.
Notice that the graph is a horizontal line while Sirius is at rest. That is only during segment A.
We stated our assumption that the origin is at the hydrant explicitly. During segments B and D time continues to increase but the position remains constant; this corresponds to zero velocity. There are five different segments of the motion, since the lines on the position-versus-time graph have different slopes between five different time periods.
To avoid his opponent's lunge, the fencer jumps backwards very quickly. He remains still for a few seconds. The fencer then begins to advance slowly on his opponent.
Velocity is given by the slope of lines on position-versus-time graphs. See Conceptual Example 2. The respective slopes matter, not how high on the graph the curves are.
The steepness of the tangent line is greatest at D. Motion to the left is indicated by a decreasing segment on the graph. Before B the object is moving right and after B it is moving left. It is amazing that we can get so much information about the velocity and even about the acceleration from a position-versus-time graph. Think about this carefully.
Notice also that the object is at rest to the left of the origin at point F. Looking at the graph, there are three time intervals where the graph has zero slope: When the slope of the lines on the graph is nonzero, the object is accelerating and therefore changing speed.
The velocity is positive while the slope of the line is negative. Since the velocity and acceleration are in opposite directions, the object is slowing down.
During segment E the slope of the line is positive which indicates positive acceleration, but the velocity is negative. Since the acceleration and velocity are in opposite directions, the object is slowing here also. Consider segment C. Here the slope of the line is negative and the velocity is negative.
The velocity and acceleration are in the same direction so the object is speeding up. The object is gaining velocity in the negative direction. In terms of the graph, positive values of velocity are above the time axis. The velocity is positive for segments A and B. The velocity must also be greater than zero. The slope of the velocity graph is the acceleration graph. This graph shows a curved position-versus-time line. Since the graph is curved the motion is not uniform. The instantaneous velocity, or the velocity at any given instant of time, is the slope of a line tangent to the graph at that point in time.
Consider the graph below, where tangents have been drawn at each labeled time. Comparing the slope of the tangents at each time in the figure above, the speed of the car is greatest at time C. Instantaneous velocity is given by the slope of a line tangent to a position-versus-time curve at a given instant of time. This is also demonstrated in Conceptual Example 2. Negative, negative; since the slope of the tangent line is negative at both 1 and 2.
The velocity of an object is given by the physical slope of the line on the position-versus-time graph. Since the graph has constant slope, the velocity is constant. We can calculate the slope by using Equation 2. The points on the line can be read to two significant figures. Since the slope is positive, the value of the position is increasing with time, as can be seen from the graph. We are asked to find the largest of four accelerations, so we compute all four from Equation 2.
A large final speed, such as in choices A and C, does not necessarily indicate a large acceleration. We are given the acceleration of the car, and need to find the stopping distance. See the pictorial representation, which includes a list of values below. An equation that relates acceleration, initial velocity, final velocity, and distance is Equation 2. We are given initial and final velocities and acceleration.
We are asked to find a displacement, so Equation 2. This is not a hard question once we remember that the displacement is the area under the velocity- versus-time graph. The scales on all three graphs are the same, so simple visual inspection will attest that Betty traveled the furthest since there is more area under her graph. The correct choice is B.
It is important to verify that the scales on the axes on all the graphs are the same before trusting such a simple visual inspection. On a related note, check the accelerations: Mentally tie this all together. The slope of the tangent to the velocity-versus-time graph gives the acceleration of each car. The slope of the tangent to the graph at the same time for Carl is larger.
See the figure below. Acceleration is given by the slope of the tangent to the curve in a velocity-versus-time graph at a given time. Both balls are in free fall neglecting air resistance once they leave the hand, and so they will have the same acceleration.
Therefore, the slopes of their velocity-versus-time graphs must be the same i. That eliminates choices B and C. Ball 1 has positive velocity on the way up, while ball 2 never goes up or has positive velocity; therefore, choice A is correct. Examine the other choices. In choice B ball 1 is going up faster and faster while ball 2 is going down faster and faster. In choice C ball 1 is going up the whole time but speeding up during the first part and slowing down during the last part; ball 2 is going down faster and faster.
There are two ways to approach this problem, and both are educational. Using algebra, first calculate the acceleration of the larger plane. The second method is graphical. Make a velocity vs. Since the area under the velocity vs. The dots from time 0 to 9 seconds indicate a direction of motion to the right. The dots are getting closer and closer.
This indicates that the object is moving to the right and slowing down. From 9 to 16 seconds, the object remains at the same position, so it has no velocity.
From 16 to 23 seconds, the object is moving to the left. Its velocity is constant since the dots are separated by identical distances. The velocity-versus-time graph that matches this motion closest is B. This can be solved with simple ratios. The answer is B. This result can be checked by actually computing the acceleration and plugging it back into the equation for the second case, but ratios are slicker and quicker.
The answer is A. This result can be checked by actually computing the acceleration, doubling it, and plugging it back into the equation for the second case, but ratios are slicker and quicker. Problems P2. The car is traveling to the left toward the origin, so its position decreases with increase in time. Let us review our sign conventions.
Position to the right of or above origin is positive, but to the left of or below origin is negative. Velocity is positive for motion to the right and for upward motion, but it is negative for motion to the left and for downward motion. The slope of the position graph is the velocity graph. The position graph has a shallow negative slope for the first 8 s, and then the slope increases. We expect the sneaking up phase to be longer than the spring phase, so this looks like a realistic situation.
To get a position from a velocity graph we count the area under the curve. There are 18 m of area above the axis and 4 m of area below.
These numbers seem reasonable; a mail carrier could back up 4 m. There are 12 m of area below the axis and 12 m of area above. Note that the slope of the position-versus-time graph at every point gives the velocity at that point.
Referring to Figure P2. As expected a positive slope gives a positive velocity and a negative slope yields a negative velocity. Assume that the ball travels in a horizontal line at a constant v x. Just under a half second is reasonable for a major league pitch. This is a short but reasonable time for a fastball to get from the mound to home plate.
Beth and Alan are moving at a constant speed, so we can calculate the time of arrival as follows: Times of the order of 7 or 8 hours are reasonable in the present problem.
Assume that Richard only speeds on the mi stretch of the interstate. We then need to compute the times that correspond to two different speeds for that given distance. Rearrange Equation 1. At the speed limit: Breaking the law to save 8. Notice how the hours as well as the miles cancel in the equations. Since each runner is running at a steady pace, they both are traveling with a constant speed. Each must travel the same distance to finish the race.
We assume they are traveling uniformly. We can calculate the time it takes each runner to finish using Equation 2. For uniform motion, velocity is given by Equation 2. Finally, subtract that distance from the 8. The faster runner finishes in 8. This leaves the slower runner 8. The slower runner will not even be in sight of the faster runner when the faster runner crosses the line. We did not need to convert hours to seconds because the hours cancelled out of the last equation.
Notice we kept 3 significant figures, as indicated by the original data. Assume vx is constant so the ratio is also constant. Setting up the ratio allows us to easily solve for the distance traveled in any given time. Assume v x is constant so the ratio is also constant. This pace does give about the right answer for the time required to run a mile for a good marathoner. The graph in Figure P2. As shown above in a , a positive slope must give a positive velocity and a negative slope must yield a negative velocity.
The distance traveled is the area under the v y graph. Four beats seems reasonable. There is some doubt that we are justified using two significant figures here. To make the estimates from the graph we need to read the slopes from the graph. Then pick a pair of points on each line to compute the rise and the run.
The speed is increasing, which is indeed what the graph tells us. These are reasonable numbers for a drag racer. Displacement is equal to the area under the velocity graph between ti and tf, and acceleration is the slope of the velocity-versus-time graph. Because the velocity was negative at first, the train was moving left. There is a turning point at 2 s. Acceleration is the rate of change of velocity. The sign conventions for position are in Figure 2. Conventions for velocity are in Figure 2.
Conventions for acceleration are in Figure 2. Since the velocity vectors are increasing in length and are pointing toward the right, the acceleration is positive. The position is always negative, but it is only differences in position that are important in calculating velocity.
Since the velocity vectors are increasing in length and are downward, the acceleration is negative. The position is always positive, but it is only differences in position that are important in calculating velocity. The origin for coordinates can be placed anywhere.
To figure the acceleration we compute the slope of the velocity graph by looking at the rise and the run for each straight line segment. Speeding up: Indeed the slope looks three time steeper in the first segment than in the second.
These are pretty large accelerations. From a velocity-versus-time graph we find the acceleration by computing the slope. We will compute the slope of each straight-line segment in the graph.
This graph is difficult to read to more than one significant figure. I did my best to read a second significant figure but there is some estimation in the second significant figure. It takes Carl Lewis almost 10 s to run m, so this graph covers only the first third of the race.
Were the graph to continue, the slope would continue to decrease until the slope is zero as he reaches his fastest cruising speed. Also, if the graph were continued out to the end of the race, the area under the curve should total m.
Use the definition of acceleration. Frogs are quite impressive! We can calculate acceleration from Equation 2. For the gazelle: A lion would have an easier time snatching a gazelle than a trout. We will then use that acceleration to compute the final position after the strike: The answer is remarkable but reasonable. The pike strikes quickly and so is able to move 0. The seconds squared cancel in the last equation.
First, we will convert units: Because the car has constant acceleration, we can use kinematic equations. A little over tenth of a mile displacement in 10 s is physically reasonable. Fleas are amazing jumpers; they can jump several times their body height—something we cannot do.
We assume constant acceleration so we can use the kinematic equations. The last of the three relates the three variables we are concerned with in part a: We do not know the time it takes the flea to reach maximum height, so we employ Equation 2. Just over 5 cm is pretty good considering the size of a flea. It is about 10—20 times the size of a typical flea. Check carefully to see that each answer ends up in the appropriate units. Assume that the acceleration during braking is constant.
There are a number of ways to approach this question. The graph will then decrease linearly and become zero at some later time t1. It must also linearly decrease to zero—and it must have the same slope because we were told the acceleration is the same in both cases. Now the crux of the matter: Carefully examine the two triangles and see that the larger one has 4 times the area of the smaller one; one way is to realize it has a base twice as large and a height twice as large, another is to mentally cut out the smaller triangle and flip and rotate it to convince yourself that four copies of it would cover the larger triangle.
Thus, the stopping distance for the 2v case is 4d. Equation 2. It demonstrates clear and versatile thinking to approach a question in multiple ways, and it gives an important check on our work. The graphical approach in this case is probably the more elegant and insightful; there is a danger that the algebraic approach can lead to blindly casting about for an equation and then plugging and chugging.
This latter mentality is to be strenuously avoided. Equations should only be used with correct conceptual understanding. So the signs work out. The answer is a reasonable 58 mph. Because the skier slows steadily, her deceleration is a constant during the glide and we can use the kinematic equations of motion under constant acceleration.
A deceleration of 2. The accelerations are same, so they cancel. It seems reasonable to need a mile for a passenger jet to take off. Because the car slows steadily, the deceleration is a constant and we can use the kinematic equations of motion under constant acceleration.
We recall that displacement is equal to area under the velocity graph between ti and tf, and acceleration is the slope of the velocity-versus-time graph. Due to the negative slope of the velocity graph between 2 s and 4 s, a negative acceleration was expected. This is a two-part problem. Then we will find the displacement as the car is brought to rest with maximum deceleration.
Keep a safe distance while driving! Do this in two parts. First compute the distance traveled during the acceleration phase and what speed it reaches. Then compute the additional distance traveled at that constant speed.