course Phy 121 I could not submit Week 4 Quiz 1 because I couldn't get the link to work. However, if there is another link to this quiz send it to me and I will submit the quiz as soon as possible. 哗fSyassignment #008
......!!!!!!!!...................................
08:35:24 Physics video clip 11 `ds on single time interval: ave. ht * width = vAve * `dt = `ds How do we calculate the displacement during a single specific time interval, given a linear v vs. t graph?
......!!!!!!!!...................................
RESPONSE --> 'ds = vAve * 'dt gives displacement.
.................................................
......!!!!!!!!...................................
08:35:50 ** STUDENT SOLUTION: ((v1+v2)/2)*(t2-t1) or area of the trapezoid created by the given time interval **
......!!!!!!!!...................................
RESPONSE --> My answer was at least partially correct.
.................................................
......!!!!!!!!...................................
08:37:43 If we know the initial and final velocities over some time interval, and if the rate at which velocity changes is constant, then how do we calculate the displacement of the object during that time interval?
......!!!!!!!!...................................
RESPONSE --> As previously stated, 'ds = vAve * 'dt. Or, the displacement of the object during a particular time interval is the area of the trapezoid under the line.
.................................................
......!!!!!!!!...................................
08:37:56 ** ((vf + v0)/2) * `dt or ((vf + v0)/2) * (t2 - t1) **
......!!!!!!!!...................................
RESPONSE --> I think I got the correct answer.
.................................................
......!!!!!!!!...................................
08:42:23 In how many different ways can we represent the calculation of the displacement over a constant-acceleration time interval?
......!!!!!!!!...................................
RESPONSE --> I would say that there are at least three different ways this calculation can be represented (at least 3) when considering the following equations and their rearrangements. vAve = 'ds / 'dt 'ds = vAve * 'dt
.................................................
......!!!!!!!!...................................
08:43:04 ** At least two: graphically as the area under the v vs. t graph and algebraically as ((vf + v0)/2) * (t2 - t1) **
......!!!!!!!!...................................
RESPONSE --> I see where I made my mistake.
.................................................
......!!!!!!!!...................................
08:51:05 Physics video clip 12 continuing ph11 How does the graphical calculation connect with our common sense about velocity, displacement and time?
......!!!!!!!!...................................
RESPONSE --> The calculation itself helps us to make sense of what we can easily deduce from the graph as well as what we already know. We can easily figure out the initial and final velocities and how that corresponds to the change in time. Displacement corresponds to the area under the line once we know the previously mentioned information. Velocity is used to signify both magnitude (numerical value) of how fast an object is moving and also the direction in which it is moving. Displacement is the change in position of an object, or how far the object is from its starting point.
.................................................
......!!!!!!!!...................................
08:51:44 ** STUDENT SOLUTION: First of all, the graph illustrates the fact that the three are related. We can graph for two of the values on the x-y plane, in order to solve for the third one. We usually graph for velocity vs. time, to find the displacement. } STUDENT QUESTION: Is it possible to find the time by setting up a graph of distance vs. velocity or velocity vs. distance? ANSWER (UNIVERSITY PHYSICS LEVEL): Excellent question. Basically if you know the velocity at every position you have completely defined the motion; however depending on the situation it might be that the best you can do is an approximation to find x vs. t (and therefore v vs. t, again in approximation). If you know the velocity at a certain position then you can predict how long it will take to get to a nearby position; knowing the velocity at that position you can predict how long it will take to get to a new nearby position. These predictions would be approximations, the accuracy of which would depend on the magnitude of the second derivative of the velocity vs. position function. They would effectively give you an estimate of position vs. clock time, from which you could also estimate velocity vs. clock time. The situation would be that you would be able to calculate for a given position the value of the rate of position change, which is the derivative of the position. If the position function is x(t) then a graph of v vs. x would be a graph of dx/dt vs. x, so that for every x you could find dx/dt. So what you would effectively have is the differential equation dx / dt = v(x). The solution of the differential equation would be your position function x(t). Depending of v(x) there might or might not be a closed-form solution of the differential equation. If not the type of numerical approximation in the paragraph before last (which is effectively Euler's Method; there are other more sophisticated and more accurate methods of approximaton) would be your only resort, and this could be done graphically. If there is a closed-form solution to the diff eq then you might or might not be able to solve the equation x = x(t) for t in terms of x, but in any case t would be given implicitly as a function of x. **
......!!!!!!!!...................................
RESPONSE --> My answer was at least partially correct.
.................................................
xSޱnh̪b assignment #008 }jhF}}Z٨ Physics I Class Notes 07-04-2007
......!!!!!!!!...................................
08:58:42 QUESTION FROM STUDENT--Please define the differnece between Fnet and Force. See if you can answer this question.
......!!!!!!!!...................................
RESPONSE --> Force is any kind of push or pull on an object. Fnet is the vector sum of all forces acting on an object. confidence assessment: 1
.................................................
......!!!!!!!!...................................
08:59:04 ** Net force is the sum of all forces acting on an object. If you're pushing your car you are exerting a force, friction is opposing you, and the let force is the sum of the two (noting that one is positive, the other negative so you end up with net force less than the force you are exerting). Your heart rate responds to the force you are exerting and the speed with which the car is moving; the accel of the car depends on the net force. **
......!!!!!!!!...................................
RESPONSE --> I think I got the correct answer. self critique assessment: 3
.................................................
......!!!!!!!!...................................
09:08:44 In terms of the equations of motion why do we expect that a * `ds is proportional to the change in v^2, and why do we then expect that the change in v^2 is proportional to Fnet `ds?
......!!!!!!!!...................................
RESPONSE --> 'ds is the distance an object travels over a given period of time. Acceleration is the change in velocity divided by the time taken to make this change. Velocity is used to signify both the magnitude (numerical value) of how fast an object is moving and also the direction in which it is moving. Because we know the three values are related a * 'ds is proportional to the change in v^2 because magnitude is a and time (when considering the area under the line between initial and final velocities on a graph); and displacement is how far the object travels. v^2 is proportional to Fnet 'ds because we know the previously mentioned variables are related. And, because we know that Fnet is the vector some of all the forces acting on an object it is only logical to come to the conclusion that Fnet would be related to v^2. confidence assessment: 1
.................................................
......!!!!!!!!...................................
09:09:43 ** It's very important in physics to be able to think in terms of proportionality. To say that y is proportional to x is to say that for some k, y = k x. That is, y is a constant multiple of x. To say that a * `ds is proportional to the change in v^2 is to say that for some k, a * `ds = k * ( change in v^2)--i.e., that a * `ds is a constant multiple of the change in v^2. In terms of the equations of motion, we know that vf^2 = v0^2 + 2 a `ds so a `ds = 1/2 (vf^2 - v0^2), which is 1/2 the change in v^2. So a `ds is a constant multiple (1/2) of the change in v^2. Formally we have a `ds = k ( change in v^2) for k = 1/2. Now since Fnet = m a we see that Fnet is proportional to a for a given mass m, and it follows that Fnet `ds = k * change in v^2, for the appropriate k (specifically for k = mass / 2. **
......!!!!!!!!...................................
RESPONSE --> I didn't explain things in terms of how the equations actually worked, but I think I was on the right track anyway. self critique assessment: 2
.................................................
......!!!!!!!!...................................
09:14:18 How do our experimental results confirm or refute this hypothesis?
......!!!!!!!!...................................
RESPONSE --> When you consider the definitions of the variables we have studied so far in class along with the equations associated as well as situations in which these mathematical phenomena can be tested using these equations, then experimental results confirm these hypothesis. confidence assessment: 1
.................................................
......!!!!!!!!...................................
09:16:00 ** We didn't actually do this part of the experiment, but on a ramp with fixed slope a `ds is simply proportional to `ds. When we measured `dt for different distances from rest down the same ramp, we were then able to determine the average and final velocities. The change in v^2 for each timing would be from 0 to vf^2. The change would therefore be just vf^2. If a `ds is proportional to the change in vf^2 then a graph of vf^2 vs. a `ds should be linear. Since a is constant we don't even need to determine it--a graph of vf^2 vs. `ds would be linear. This would confirm the hypothesis. **
......!!!!!!!!...................................
RESPONSE --> I see where I made my mistake, but I knew somehow the equations would prove the hypothesis. self critique assessment: 2
.................................................
Ζڍeޜ assignment #008 }jhF}}Z٨ Physics I Class Notes 07-04-2007
......!!!!!!!!...................................
11:21:10 What is the meaning of the x intercept of the graph of force vs. slope?
......!!!!!!!!...................................
RESPONSE --> This information gives displacement when a certain amount of force is applied.
.................................................
......!!!!!!!!...................................
11:22:09 ** On this graph the x axis represents slope, the y axis represents acceleration. The x axis point occurs when y = 0, so this point represents a slope for which acceleration is zero. Assuming that 'down the incline' is the positive direction, everywhere to the right of this point the acceleration is positive and everywhere to the left is negative. The x intercept therefore represents the maximum slope for which friction is great enough to prevent the ball from accelerating down the slope. Since friction is resisting the weight component down the incline, the x intercept represents the slope when friction and the weight component tending to accelerate the cart down the slope are exactly in balance, equal and opposite. **
......!!!!!!!!...................................
RESPONSE --> I am not sure where I went wrong, but this solution makes sense.
.................................................
......!!!!!!!!...................................
11:24:38 When we obtain a linear relationship between force and acceleration, is it plausible that the constant term in the equation is, within experimental uncertainty, zero?
......!!!!!!!!...................................
RESPONSE --> Yes, the object won't accelerate unless a force is acting upon it, or it will accelerate at a rate of zero and maintain a constant velocity as long as no other forces act on the object.
.................................................
......!!!!!!!!...................................
11:25:18 ** The constant term in the equation will be zero only if the graph passes through the origin. This would imply that any nonzero force will produce a nonzero acceleration, and that for example there is either no friction present or friction has been compensated for in some way. **
......!!!!!!!!...................................
RESPONSE --> I see where I made my mistake. I was thinking of the equations instead of what the graph actually shows.
.................................................
......!!!!!!!!...................................
11:26:41 For the force vs. acceleration relationship for the car, why should we expect that the acceleration corresponding to a net force equal to the car's weight is the acceleration of gravity?
......!!!!!!!!...................................
RESPONSE --> Because the car is accelerating on a horizontal surface, and assuming that friction and wind resistance are negligible then gravity would be the most significant force acting on the car.
.................................................
......!!!!!!!!...................................
11:27:04 ** When the car is dropped and allowed to fall freely the net force acting on it is its weight, which is the force exerted on its mass by gravity. So the acceleration will be that of gravity. **
......!!!!!!!!...................................
RESPONSE --> My answer was at least partially correct.
.................................................
HǾRHў assignment #008 008. Using the Acceleration of Gravity Physics I 07-04-2007
......!!!!!!!!...................................
09:19:54 `q001. The accepted value of the acceleration of gravity is approximately 980 cm/s^2 or 9.8 m/s^2. This will be the acceleration, accurate at most places within 1 cm/s^2, of any object which falls freely, that is without the interference of any other force, near the surface of the Earth. If you were to step off of a table and were to fall 1 meter without hitting anything, you would very nearly approximate a freely falling object. How fast would you be traveling when you reached the ground?
......!!!!!!!!...................................
RESPONSE --> Force equals mass times acceleration. My mass along with the acceleration, which in this case would be gravity (9.8 m/s^2) if no other forces were acting on me (such as wind resistance, friction, etc.), would provide how much force. confidence assessment: 1
.................................................
......!!!!!!!!...................................
09:21:18 You would have an initial vertical velocity of 0, and would accelerate at 9.8 m/s^2 in the same direction as your 1 meter vertical displacement. You would also have a slight horizontal velocity (you don't step off of a table without moving a bit in the horizontal direction, and you would very likely maintain a small horizontal velocity as you fell), but this would have no effect on your vertical motion. So your vertical velocity is a uniform acceleration with v0 = 0, `ds = 1 meter and a = 9.8 m/s^2. The equation vf^2 = v0^2 + 2 a `ds contains the three known variables and can therefore be used to find the desired final velocity. We obtain vf = +- `sqrt( v0^2 + 2 a `ds) = +- `sqrt ( 0^2 + 2 * 9.8 m/s^2 * 1 m)= +- `sqrt ( 19.6 m^2 / s^2) = +- 4.4 m/s, approx. Since the acceleration and displacement were in the direction chosen as positive, we conclude that the final velocity will be in the same direction and we choose the solution vf = +4.4 m/s.
......!!!!!!!!...................................
RESPONSE --> I was at least partially correct, and I understand this solution. self critique assessment: 2
.................................................
......!!!!!!!!...................................
09:58:42 `q002. If you jump vertically upward, leaving the ground with a vertical velocity of 3 m/s, how high will you be at the highest point of your jump? Note that as soon as you leave the ground, you are under the influence of only the gravitational force. All the forces that you exerted with your legs and other parts of your body to attain the 3 m/s velocity have done their work and are no longer acting on you. All you have to show for it is that 3 m/s velocity. So as soon as you leave the ground, you begin experiencing an acceleration of 9.8 m/s^2 in the downward direction. Now again, how high will you be at the highest point of your jump?
......!!!!!!!!...................................
RESPONSE --> I'm not entirely sure of how I am to approach this, but to get the units of both of these numbers so we can work with them I took the square root of both values to the the square of 9.8 m/s^2 to cancel. That left me with 1.73 and 3.13, and the difference between these values is -1.4 m/s which is the rate of force acting against upward acceleration. 3 - 1.4 = 1.6 m/s which is actually the rate at which you are accelerating upward after you figure in the force of gravity. The average person doesn't usually get more than a couple of feet off the ground. Therefore: 24 in. * (1 cm / 2.54 in.) * ( 10^-2 m / 1 cm) = .0945 m confidence assessment: 1
.................................................
......!!!!!!!!...................................
10:00:29 From the instant the leave the ground until the instant you reach your highest point, you have an acceleration of 9.8 m/s^2 in the downward direction. Since you are jumping upward, and since we can take our choice of whether upward or downward is the positive direction, we choose the upward direction as positive. You might have chosen the downward direction, and we will see in a moment how you should have proceeded after doing so. For now, using the upward direction as positive, we see that you have an initial velocity of v0 = + 3 m/s and an acceleration of a = -9.8 m/s^2. In order to use any of the equations of motion, each of which involves four variables, you should have the values of three variables. So far you only have two, v0 and a. {}What other variable might you know? If you think about it, you will notice that when objects tossed in the air reach their highest point they stop for an instant before falling back down. That is precisely what will happen to you. At the highest point your velocity will be 0. Since the highest point is the last point we are considering, we see that for your motion from the ground to the highest point, vf = 0. Therefore we are modeling a uniform acceleration situation with v0 = +3 m/s, a = -9.8 m/s^2 and vf = 0. We wish to find the displacement `ds. Unfortunately none of the equations of uniformly accelerated motion contain the four variables v0, a, vf and `ds. This situation can be easily reasoned out from an understanding of the basic quantities. We can find the change in velocity to be -3 meters/second; since the acceleration is equal to the change in velocity divided by the time interval we quickly determine that the time interval is equal to the change in velocity divided by the acceleration, which is `dt = -3 m/s / (-9.8 m/s^2) = .3 sec, approx.; then we multiply the .3 second time interval by the 1.5 m/s average velocity to obtain `ds = .45 meters. However if we wish to use the equations, we can begin with the equation vf = v0 + a `dt and solve to find `dt = (vf - v0) / a = (0 - 3 m/s) / (-9.8 m/s^2) = .3 sec. We can then use the equation `ds = (vf + v0) / 2 * `dt = (3 m/s + 0 m/s) / 2 * .3 sec = .45 m. This solution closely parallels and is completely equivalent to the direct reasoning process, and shows that and initial velocity of 3 meters/second should carry a jumper to a vertical height of .45 meters, approximately 18 inches. This is a fairly average vertical jump. If the negative direction had been chosen as positive then we would have a = +9.8 m/s^2, v0 = -3 m/s^2 (v0 is be in the direction opposite the acceleration so if acceleration is positive then initial velocity is negative) and again vf = 0 m/s (0 m/s is the same whether going up or down). The steps of the solution will be the same and the same result will be obtained, except that `ds will be -.45 m--a negative displacement, but where the positive direction is down. That is we move .45 m in the direction opposite to positive, meaning we move .45 meters upward.
......!!!!!!!!...................................
RESPONSE --> I see where I made my mistake. I was on the right track to thinking in the way I should have been anyway. self critique assessment: 2
.................................................
......!!!!!!!!...................................
10:15:25 `q003. If you roll a ball along a horizontal table so that it rolls off the edge of the table at a velocity of 3 m/s, the ball will continue traveling in the horizontal direction without changing its velocity appreciably, and at the same time will fall to the floor in the same time as it would had it been simply dropped from the edge of the table. If the vertical distance from the edge of the table to the floor is .9 meters, then how far will the ball travel in the horizontal direction as it falls?
......!!!!!!!!...................................
RESPONSE --> Using the equations of motion, we are looking for 'dt. We know v0, vf, and a so we can use the following equations to find what we need to know. vf = v0 + a * 'dt 0 = 0 + 3 'dt 3 sec = 'dt
.................................................
......!!!!!!!!...................................
10:16:08 A ball dropped from rest at a height of .9 meters will fall to the ground with a uniform vertical acceleration of 9.8 m/s^2 downward. Selecting the downward direction as positive we have `ds = .9 meters, a = 9.8 m/s^2 and v0 = 0. Using the equation `ds = v0 `dt + .5 a `dt^2 we see that v0 = 0 simplifies the equation to `ds = .5 a `dt^2, so `dt = `sqrt( 2 `ds / a) = `sqrt(2 * .9 m / (9.8 m/s^2) ) = .42 sec, approx.. Since the ball rolls off the edge of the table with only a horizontal velocity, its initial vertical velocity is still zero and it still falls to the floor in .42 seconds. Since its horizontal velocity remains at 3 m/s, it travels through a displacement of 3 m/s * .42 sec = 1.26 meters in this time.
......!!!!!!!!...................................
RESPONSE --> I'm not sure where I went wrong here, comments or suggestions would be helpful. self critique assessment: 2
................................................."