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20:53:11 What do we mean by velocity?
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RESPONSE --> Velocity is the rate and direction that an object travels.
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20:53:55 ** STUDENT RESPONSE:Velocity is the speed and direction an object is moving. INSTRUCTOR COMMENT: Good. More succinctly and precisely velocity is the rate at which position is changing. obtained by dividing change in position by change in clock time **
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RESPONSE --> Ok.
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20:55:14 How can we determine the velocity of a ball rolling down an incline?
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RESPONSE --> To determine the velocity of a ball rolling down an incline, we divide the distance that the balls traveled by the time it takes the ball to reach the end of the incline.
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20:57:13 ** We divide displacement by the time interval to get average velocity. If the ball happens to be starting from rest and acceleration is uniform we can also conclude that the final velocity attained on the incline is double the average velocity. **
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RESPONSE --> Ok. I must of missed the question asking for the average velocity instead of just velocity alone. I know the difference between the two (average uses the changes in distance and time whereas velocity does not).
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21:05:08 We anticipate from our experience that a ball traveling down a greater incline, starting from rest, will experience a greater change in velocity. How can we determine whether the velocity actually changes, and whether the velocity increases in the manner that we expect?
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RESPONSE --> I think that if the ramp is not at a constant incline, then we would know that the velocity will not be constant as the ball rolls down it. The velocity will increase if the ramp is at a higher incline at the beginning, but velocity will decrease if the ramp is inclined higher at the end.
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21:07:02 ** We divide displacement by the time interval to get average velocity. We time the ball down one incline, then down the other and determine average velocity for each.
We then infer that since both balls started from rest, the greater average velocity implies a greater change in velocity. **......!!!!!!!!...................................
RESPONSE --> Ok I see what the question was pertaining to. The greater the change in velocity, the greater the average velocity will be. So a car that goes from 0-60 in 5 seconds has a greater velocity of one that goes 0-40 in 5 seconds. That seems like common sense...
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21:09:19 How could we determine the velocity of the ball at a specific point? The specific points are measured for distance and the ball is timed when it reaches these specific points. The distance is then divided by the time.
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RESPONSE --> To get the velocity of the ball at a certain point, we would need to figure out the average velocity. Take the difference of the distance and the time at the specific point and divide and it'll give you the average cm/s/s of the ball (assuming those are the units needed).
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21:11:16 ** Short answer: The question concerned one specific point. We can't really measure this precisely. The best we can do is use two points close together near the point we are interested in, but not so close we can't measure the time accurately enough to trust our result.
More detailed answer: The question really asks how we determine the velocity at a given point, for an object in the real world. Assuming that the velocity is always changing, how can we ever know the velocity at an instant? This involves a limiting process, thinking of shorter and shorter time intervals and shorter and shorter position changes. If we know the velocity function, or if we can accurately infer the velocity function from our data, then the velocity of a ball at a specific point is obtained by finding the slope of the tangent line of the position vs. clock time graph at that point, which calculus-literate students will recognize as the derivative of the velocity function. **......!!!!!!!!...................................
RESPONSE --> The short answer is exactly what I was thinking though I'm not sure if I worded it as clearly. As for the more detailed answer, it somewhat makes sense, but I'm not exactly calculus-literate, even if I did take years of it.
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21:19:52 How do we determine the rate at which the velocity changes? How can we understand the concept of the rate at which velocity changes?
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RESPONSE --> Use what is called the slope triangle to figure out the rate at which the velocity changes. Figure out the slope of the curve in two different places and use them to find the velocity. The closer together that the two points get to each other, eventually they'll reach the tangent line.
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21:21:03 ** We find the change in velocity then divide by the change in the clock time. Any rate consists of the change in one quantity divided by the change in another. **
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RESPONSE --> Ok so I went overboard and bypassed the most simple definition available. That and it's been so long that I've forgotten the definition of a rate already.
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21:26:06 It is essential to understand what a trapezoid on a v vs. t graph represents. Give the meaning of the rise and run between two points, and the meaning of the area of a trapezoid defined by a v vs. t graph.
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RESPONSE --> The rise and the run between two points on a graph is what gives you the slope. As for the meaning of the area of a trapezoid, I'm not sure.
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21:32:04 ** Since the rise represents the change in velocity and the run represents the change in clock time, slope represents `dv / `dt = vAve, the average velocity over the corresponding time interval.
Since the average altitude represents the average velocity and the width of the trapezoid represents the time interval the area of the trapezoid represents vAve * `dt, which is the displacement `ds. **......!!!!!!!!...................................
RESPONSE --> Ok I remember the book mentioning the displacement being what is needed for the average velocity. I just don't remember it mentioning the trapezoid area, only that of a triangle between two points on the graph. The way the response is written, I'm hoping I'm correct when I say that the slope multiplied by the change in time is what gives the displacement variable. I tend to get confused with the vAve and 'dt symbols and forget what they stand for.
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22:17:23 What does the graph of position vs. clock time look like for constant-acceleration motion?
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RESPONSE --> I have no idea what the graph of position v/s clock time will look like. I have the dvds for the course and the class notes and all the other files seem to be real scattered and not labled clearly and I could not find anything on acceleration in them.
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22:20:03 ** For constant positive acceleration velocity is increasing. The greater the velocity the steeper the position vs. clock time graph. So increasing velocity would be associated with a position vs. clock time graph which is increasing at an increasing rate.
The reason velocity is the slope of the position vs. clock time graph is that the rise between two points of the position vs. clock time graph is change in position, `ds, and run is change in clock time, `dt. Slope therefore represents `ds / `dt, which is velocity. Other shapes are possible, depending on whether initial velocity and acceleration are positive, negative or zero. For example if acceleration was negative and initial velocity positive we could have a graph that's increasing at a decreasing rate. Negative initial velocity and positive acceleration could give us a graph that's decreasing at a decreasing rate. **......!!!!!!!!...................................
RESPONSE --> Ok I think I understand this.
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22:26:20 How can we obtain a graph of velocity vs. clock time from a position vs. clock time graph?
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RESPONSE --> Once again, I could not find the notes for this on the dvds, so I'm just guessing off the last question that the steeper the position v/s clock time graph, the greater the slope or velocity will be. I'm really kind of confused with this having not found the files yet. I've spent hours watching video clips looking for the file with this information and haven't found it.
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22:28:41 ** We can find the slope of the position vs. clock time graph at a series of clock times, which will give us the velocities at those clock times. We can put this information into a velocity vs. clock time table then plot the velocities vs. clock time as a 'guidepost points', and fill in the connecting curve in such a way as to be consistent with the trend of the slopes of the position vs. clock time graph.
COMMON MISCONCEPTION: To get velocity vs. clock time find average velocity, which is position (m) divided by time (s). Plot these points of vAvg on the velocity vs. time graph. INSTRUCTOR RESPONSE: Ave velocity is change in position divided by change in clock time. It is not position divided by time. Position can be measured from any reference point, which would affect a position/time result, but which would not affect change in position/time. Graphically velocity is the slope of the position vs. clock time graph. If it was just position divided by time, it would be the slope of a line from the origin to the graph point. **......!!!!!!!!...................................
RESPONSE --> Ok I kind of understand this. I'm not really sure what about it doesn't make sense so I can't really critique my answer/thoughts.
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22:35:01 How can we obtain a graph of position vs. clock time from a velocity vs. clock time graph
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RESPONSE --> Once again, I have no idea. Maybe find the displacement or the average velocity for a specific position on the graph, but I'm just guessing again.
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22:39:04 ** We can divide the graph of v vs. t into small strips, each forming an approximate trapezoid. The area of each strip will represent ave vel * time interval and will therefore represent the change in position during that time interval. Starting from the initial clock time and position on the position vs. clock time graph, we add each subsequent time increment to the clock time and the corresponding position change to the position to get our new position.
When the graph is constructed the slopes of the position vs. clock time graph will indicate the corresponding velocities on the v vs. t graph. **......!!!!!!!!...................................
RESPONSE --> Ok this makes more sense than the previous questions because it was easier to visualize. I did see a section of class notes that seemed to show what this explanation is describing and I think I was slightly correct when I said to find the average velocity for a specific point whereas the answer said to split the entire graph up and find the average velocity for numerous different sections and add them together to get the new position.
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13:07:30 How can we obtain a graph of acceleration vs. clock time from a velocity vs. clock time graph?
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RESPONSE --> I'm pretty sure if you take the slope of the velocity, it represents the acceleration, and then you could use the slopes from different points on the graph to make a new graph of the acceleration. I found some notes online and this is what I got from them but I'm not sure if it's right...
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13:08:50 ** Accel is the rate of change of velocity, represented by the slope of the v vs. t graph. So we would plot the slope of the v vs. t graph vs. t, in much the same way as we plotted slopes of the position vs. clock time graph to get the v vs. t graph.
}University Physics Students note: Acceleration is the derivative of the velocity. COMMON MISCONCEPTION: Take speed/ time to find the acceleration per second. The form an acceleration v. time graph and draw a straight line out from the number calculated for acceleration above. INSTRUCTOR RESPONSE: Ave acceleration is change in velocity divided by change in clock time. (note that this is different from velocity divided by time--we must use changes in velocity and clock time). (Advanced note: Velocity is always measured with respect to some reference frame, and the velocity of the reference frame itself affects a velocity/time result, but which would not affect change in velocity/time). Graphically acceleration is the slope of the velocity vs. clock time graph. If it was velocity divided by time, it would be the slope of a line from the origin to the graph point. **......!!!!!!!!...................................
RESPONSE --> Phew, I finally found the right notes to use for this program.
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13:16:59 How can we obtain a graph of velocity vs. clock time from an acceleration vs. clock time graph
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RESPONSE --> I think you subtract the final velocity and the velocity to get the acceleration.
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13:18:48 ** STUDENT RESPONSE: Take your acceleration and multiply by time to find the change in velocity. Start with initial velocity and graph your velocity by increasing initial velocity by the slope, or change in velocity. INSTRUCTOR COMMENT: Good. More precisely we can approximate change in velocity during a given time interval by finding the approximate area under the acceleration vs. clock time graph for the interval. We can then add each change in velocity to the existing velocity, constructing the velocity vs. clock time graph interval by interval.
A velocity vs. clock time graph has slopes which are equal at every point to the vertical coordinate of the acceleration vs. clock time graph. University Physics students note: These two statements are equivalent, and the reason they are is at the heart of the Fundamental Theorem of Calculus. **......!!!!!!!!...................................
RESPONSE --> Ok. I'm still not 100% clear on this but I think I'm beginning to understand more of it than before now that I've found the notes on the dvds.
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GޟcS嬰}b assignment #001 ~ȪsJ콥 Physics I Vid Clips 09-24-2005
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19:54:16 Physics video clip 01: A ball rolls down a straight inclined ramp. It is the velocity the ball constant? Is the velocity increasing? Is the velocity decreasing?
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RESPONSE --> The velocity of the ball is constant.
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19:57:21 ** It appears obvious, from common experience and from direct observation, that the velocity of the ball increases.
A graph of position vs. clock time would be increasing, indicating that the ball is moving forward. Since the velocity increases the position increases at an increasing rate, so the graph increases at an increasing rate. **......!!!!!!!!...................................
RESPONSE --> Oh ok I see how I misunderstood that video clip. I figured since the motion of the ball didn't suddenly speed up or slow down during the duration on the ramp, then the velocity remained constant. But since the ramp caused it to roll in the first place, then the velocity increased.
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20:00:55 If the ball had a speedometer we could tell. What could we measure to determine whether the velocity of the ball is increase or decreasing?
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RESPONSE --> You could measure different points on the graph to get the average rate of velocity change. That I think is the change in distance divided by the time interval.
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20:02:33 ** STUDENT RESPONSE: By measuring distance and time we could calculate velocity.
INSTRUCTOR COMMENTS: The ball could be speeding up or slowing down--all you could get from the calculation you suggest is the average velocity. You could measure the time to travel the first half and the time to travel the second half of the ramp; if the latter is less then we would tend to confirm increasing velocity (though those are still average velocities and we wouldn't get certain proof that the velocity was always increasing). You would need at least two velocities to tell whether velocity is increasing or decreasing. So you would need two sets of distance and time measurements. **......!!!!!!!!...................................
RESPONSE --> I didn't state clearly that two measurements were needed to be compared to each other, but I was in fact picturing it being done that way. Anything over one measurement would work out fine.
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20:03:17 What is the shape of the velocity vs. clock time graph for the motion of the ball?
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RESPONSE --> The curve is increasing at an increasing rate.
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20:03:48 ** If the ramp has an increasing slope, the velocity would increase at an increasing rate and the graph would curve upward, increasing at an increasing rate. If the ramp has a decreasing slope, like a hill that gradually levels off, the graph would be increasing but at a decreasing rate. On a straight incline it turns out that the graph would be linear, increasing at a constant rate, though you aren't expected to know this at this point. All of these answers assume an absence of significant frictional forces such as air resistance. **
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RESPONSE --> ok
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20:10:54 A ball rolls down ramp which curves upward at the starting end and otherwise rests on a level table. What is the shape of the velocity vs. clock time graph for the motion of the ball?
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RESPONSE --> I think the ball is increasing at a decreasing rate. Since it starts off quicker than when it finishes, I'm guessing that the speed will decrease towards the end, but I wouldn't be surprised if I was wrong on this also.
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20:11:22 ** While on the curved end the ball will be speeding up, and the graph will therefore rise. By the time the ball gets to the level part the velocity will no longer be increasing and the graph will level off; because of friction the graph will actually decrease a bit, along a straight line. As long as the ball is on the ramp the graph will continue on this line until it reaches zero, indicating that the ball eventually stops. In the ideal frictionless situation on an infinite ramp the line just remains level forever. **
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RESPONSE --> ok
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20:13:47 For the ball on the straight incline, we would certainly agree that the ball's velocity is increasing. Is the velocity increasing at a constant, an increasing, or a decreasing rate? What does the graph of velocity vs. clock time look like?
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RESPONSE --> The ball is increasing at a constant rate since the ramp remains the same throughout. The graph is simply a straight line.
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20:39:51 ** It turns out that on a straight incline the velocity increases at a constant rate, so the graph is a straight line which increases from left to right.
Note for future reference that a ball on a constant incline will tend to have a straight-line v vs. t graph; if the ball was on a curved ramp its velocity vs. clock time graph would not be straight, but would deviate from straightness depending on the nature of the curvature (e.g., slope decreasing at increasing rate implies v vs. t graph increasing at increasing rate).**......!!!!!!!!...................................
RESPONSE --> ok
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`EDZz옘JŀH| Student Name: assignment #001 001. Areas
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14:06:59 `q001. There are 11 questions and 7 summary questions in this assignment.
What is the area of a rectangle whose dimensions are 4 m by 3 meters.......!!!!!!!!...................................
RESPONSE --> A = lw is the equation for the area of a rectangle. A = 3*4 = 12m^2 is the area.
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14:07:26 A 4 m by 3 m rectangle can be divided into 3 rows of 4 squares, each 1 meter on a side. This makes 3 * 4 = 12 such squares. Each 1 meter square has an area of 1 square meter, or 1 m^2. The total area of the rectangle is therefore 12 square meters, or 12 m^2.
The formula for the area of a rectangle is A = L * W, where L is the length and W the width of the rectangle. Applying this formula to the present problem we obtain area A = L * W = 4 m * 3 m = (4 * 3) ( m * m ) = 12 m^2. Note the use of the unit m, standing for meters, in the entire calculation. Note that m * m = m^2.......!!!!!!!!...................................
RESPONSE --> Ok
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14:10:37 `q002. What is the area of a right triangle whose legs are 4.0 meters and 3.0 meters?
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RESPONSE --> The equation for a right triangle is C^2 = a^2 + b^2 So C^2 = 4^2 + 3^2 C^2 = 16 + 9 = 25 Square root of 25 = 5m is the area of the right triangle
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14:12:38 A right triangle can be joined along its hypotenuse with another identical right triangle to form a rectangle. In this case the rectangle would have dimensions 4.0 meters by 3.0 meters, and would be divided by any diagonal into two identical right triangles with legs of 4.0 meters and 3.0 meters.
The rectangle will have area A = L * W = 4.0 m * 3.0 m = 12 m^2, as explained in the preceding problem. Each of the two right triangles, since they are identical, will therefore have half this area, or 1/2 * 12 m^2 = 6.0 m^2. The formula for the area of a right triangle with base b and altitude h is A = 1/2 * b * h.......!!!!!!!!...................................
RESPONSE --> I used the wrong equation. Instead of using the equation to figure out the area, I used the equation to figure out the hypotenuse.
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14:14:01 `q003. What is the area of a parallelogram whose base is 5.0 meters and whose altitude is 2.0 meters?
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RESPONSE --> The equation for the area of a parallelogram is A = bh. A = 5m * 2m = 10m^2
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14:14:22 A parallelogram is easily rearranged into a rectangle by 'cutting off' the protruding end, turning that portion upside down and joining it to the other end. Hopefully you are familiar with this construction. In any case the resulting rectangle has sides equal to the base and the altitude so its area is A = b * h.
The present rectangle has area A = 5.0 m * 2.0 m = 10 m^2.......!!!!!!!!...................................
RESPONSE --> ok
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14:15:47 `q004. What is the area of a triangle whose base is 5.0 cm and whose altitude is 2.0 cm?
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RESPONSE --> The equation needed is again, A = 1/2 bh. A = 1/2 (5m*2m) A = 1/2 *10m^2 A = 5m^2
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14:16:24 It is possible to join any triangle with an identical copy of itself to construct a parallelogram whose base and altitude are equal to the base and altitude of the triangle. The area of the parallelogram is A = b * h, so the area of each of the two identical triangles formed by 'cutting' the parallelogram about the approriate diagonal is A = 1/2 * b * h. The area of the present triangle is therefore A = 1/2 * 5.0 cm * 2.0 cm = 1/2 * 10 cm^2 = 5.0 cm^2.
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RESPONSE --> I got my units wrong but the equation is correct this time around.
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14:26:40 `q005. What is the area of a trapezoid with a width of 4.0 km and average altitude of 5.0 km?
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RESPONSE --> the equation for the area of a trapezoid is A = h/2 (a+b).
Though I'm not sure since there's three variables needed for the equation and only two given in the problem, I'm guessing that the calculation is A = 5/2 (4). A = 10km^2..................................................
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14:31:45 Any trapezoid can be reconstructed to form a rectangle whose width is equal to that of the trapezoid and whose altitude is equal to the average of the two altitudes of the trapezoid. The area of the rectangle, and therefore the trapezoid, is therefore A = base * average altitude. In the present case this area is A = 4.0 km * 5.0 km = 20 km^2.
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RESPONSE --> My calculus book gave me a different equation for a trapezoid, which is where I got confused when I didn't have the variables that I needed. The class notes I watched called a triangle a trapezoid and now this says that a rectangle is too... It's been 7 years since I took trig so I'm not surprised that this doesn't really settle well with me, but I am wondering why and how a trapezoid is both a triangle and a rectangle...
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14:44:17 `q006. What is the area of a trapezoid whose width is 4 cm in whose altitudes are 3.0 cm and 8.0 cm?
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RESPONSE --> Since the average of the two altitudes of a trapezoid equals the altitude of a rectangle. So I averaged 3cm + 8cm = 11cm 11/2 = 5.5cm Now I get the area of a rectangle A = lw. A = 5.5 * 4 = 22cm^2
I also figured out with the equation I used from earlier that you could also do A = 4/2 (3+8) = 2(11) = 22cm^2..................................................
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14:44:26 The area is equal to the product of the width and the average altitude. Average altitude is (3 cm + 8 cm) / 2 = 5.5 cm so the area of the trapezoid is A = 4 cm * 5.5 cm = 22 cm^2.
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RESPONSE --> Ok.
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14:46:54 `q007. What is the area of a circle whose radius is 3.00 cm?
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RESPONSE --> The equation for the area of a circle is A = pi r^2 A = 3.14 (3^2) A = 28.3cm^2
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14:48:49 The area of a circle is A = pi * r^2, where r is the radius. Thus
A = pi * (3 cm)^2 = 9 pi cm^2. Note that the units are cm^2, since the cm unit is part r, which is squared. The expression 9 pi cm^2 is exact. Any decimal equivalent is an approximation. Using the 3-significant-figure approximation pi = 3.14 we find that the approximate area is A = 9 pi cm^2 = 9 * 3.14 cm^2 = 28.26 cm^2, which we round to 28.3 cm^2 to match the number of significant figures in the given radius. Be careful not to confuse the formula A = pi r^2, which gives area in square units, with the formula C = 2 pi r for the circumference. The latter gives a result which is in units of radius, rather than square units. Area is measured in square units; if you get an answer which is not in square units this tips you off to the fact that you've made an error somewhere.......!!!!!!!!...................................
RESPONSE --> ok
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14:50:17 `q008. What is the circumference of a circle whose radius is exactly 3 cm?
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RESPONSE --> The equation for the circumference of a circle is A = 2 pi r. A = 2 pi (3) = 18.8cm
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14:50:31 The circumference of this circle is
C = 2 pi r = 2 pi * 3 cm = 6 pi cm. This is the exact area. An approximation to 3 significant figures is 6 * 3.14 cm = 18.84 cm. Note that circumference is measured in the same units as radius, in this case cm, and not in cm^2. If your calculation gives you cm^2 then you know you've done something wrong.......!!!!!!!!...................................
RESPONSE --> ok
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14:53:58 `q009. What is the area of a circle whose diameter is exactly 12 meters?
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RESPONSE --> Since the diameter is given, there's two routes you could take to get the area. Since the radius is half of the diameter, you could divide 12/2 = 6 and use the A = pi r^2 equation to get 113m^2.
You could also use the equation A = (pi d^2) / 4 A = 452.4 / 4 = 113m^2..................................................
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14:54:19 The area of a circle is A = pi r^2, where r is the radius. The radius of this circle is half the 12 m diameter, or 6 m. So the area is
A = pi ( 6 m )^2 = 36 pi m^2. This result can be approximated to any desired accuracy by using a sufficient number of significant figures in our approximation of pi. For example using the 5-significant-figure approximation pi = 3.1416 we obtain A = 36 m^2 * 3.1416 = 113.09 m^2.......!!!!!!!!...................................
RESPONSE --> ok
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14:57:16 `q010. What is the area of a circle whose circumference is 14 `pi meters?
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RESPONSE --> Though I'm not sure what to do here, I'm guessing by previous explanations that multiplying 14 * pi will give the area of 43.98m^2.
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14:59:15 We know that A = pi r^2. We can find the area if we know the radius r. We therefore attempt to use the given information to find r.
We know that circumference and radius are related by C = 2 pi r. Solving for r we obtain r = C / (2 pi). In this case we find that r = 14 pi m / (2 pi) = (14/2) * (pi/pi) m = 7 * 1 m = 7 m. We use this to find the area A = pi * (7 m)^2 = 49 pi m^2.......!!!!!!!!...................................
RESPONSE --> Ok I should have been able to figure that out but I didn't even think of rearranging the circumference equation to get the radius.
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15:02:26 `q011. What is the radius of circle whose area is 78 square meters?
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RESPONSE --> I need to rearrange the A = pi r^2 equation to solve for r. So 78 / pi = r^2. Then take the square root of 78 / pi to get r = 4.98m.
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15:02:37 Knowing that A = pi r^2 we solve for r. We first divide both sides by pi to obtain A / pi = r^2. We then reverse the sides and take the square root of both sides, obtaining r = sqrt( A / pi ).
Note that strictly speaking the solution to r^2 = A / pi is r = +-sqrt( A / pi ), meaning + sqrt( A / pi) or - sqrt(A / pi). However knowing that r and A are both positive quantities, we can reject the negative solution. Now we substitute A = 78 m^2 to obtain r = sqrt( 78 m^2 / pi) = sqrt(78 / pi) m.{} Approximating this quantity to 2 significant figures we obtain r = 5.0 m.......!!!!!!!!...................................
RESPONSE -->
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15:04:51 `q012. Summary Question 1: How do we visualize the area of a rectangle?
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RESPONSE --> I simply drew a rectangle and labeled where the length and width belonged. I'm not quite sure how to visualize one.
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15:05:51 We visualize the rectangle being covered by rows of 1-unit squares. We multiply the number of squares in a row by the number of rows. So the area is A = L * W.
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RESPONSE --> ok I see what the question was asking for. I don't fill any of the diagrams in with unit boxes but I see how it would work without knowing the equation in some cases.
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15:07:30 `q013. Summary Question 2: How do we visualize the area of a right triangle?
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RESPONSE --> Drawing out a right triangle shows that it can be doubled and made into a rectangle. So in order to get the area of a triangle, all that is needed is to divide the area of a rectangle by 2. I guess drawing unit boxes may work also but it's not something I'm prone to do.
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15:07:41 We visualize two identical right triangles being joined along their common hypotenuse to form a rectangle whose length is equal to the base of the triangle and whose width is equal to the altitude of the triangle. The area of the rectangle is b * h, so the area of each triangle is 1/2 * b * h.
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RESPONSE --> ok
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15:08:21 `q014. Summary Question 3: How do we calculate the area of a parallelogram?
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RESPONSE --> Since a parallelogram can be rearranged to make a rectangle, then the same equation for the area of a rectangle can be applied. A = bh.
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15:08:30 The area of a parallelogram is equal to the product of its base and its altitude. The altitude is measured perpendicular to the base.
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RESPONSE --> ok
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15:10:03 `q015. Summary Question 4: How do we calculate the area of a trapezoid?
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RESPONSE --> Though I'm not 100% sure as to how a trapezoid can be rearranged to make a rectangle, the same equation can be applied. A = lw. I think that's right.
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15:10:58 We think of the trapezoid being oriented so that its two parallel sides are vertical, and we multiply the average altitude by the width.
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RESPONSE --> Ok I forgot about averaging the two altitudes. That's only needed if both altitudes are given, otherwise I think my original answer can still work.
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15:11:55 `q016. Summary Question 5: How do we calculate the area of a circle?
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RESPONSE --> The area of a circle is calculated by multiplying pi by r^2.
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15:12:00 We use the formula A = pi r^2, where r is the radius of the circle.
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RESPONSE --> ok
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15:13:21 `q017. Summary Question 6: How do we calculate the circumference of a circle? How can we easily avoid confusing this formula with that for the area of the circle?
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RESPONSE --> The circumference of a circle is C = 2 pi r. Confusion is avoided by looking at the units in the answer. If the units come out squared (i.e. cm^2), then that means that something was done wrong. If the units come out as just plain cm, then that is the circumference.
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15:13:26 We use the formula C = 2 pi r. The formula for the area involves r^2, which will give us squared units of the radius. Circumference is not measured in squared units.
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RESPONSE --> ok
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15:15:16 `q018. Explain how you have organized your knowledge of the principles illustrated by the exercises in this assignment.
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RESPONSE --> I did cheat and use the physics book to get some of the equations from (mainly the parallelogram and trapezoid). It was also good to have a reminder of how to go about rearranging the equations to get answers. I'm surprised that that concept slipped my mind but hopefully it won't in the future.
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15:15:22 This ends the first assignment.
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RESPONSE --> ok
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rxy~x Student Name: assignment #001
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21:05:59 This exercise was to have included a simulation of the pendulum as a timer. However due to delays in the startup of the course this exercise will not be included this semester. Your real first assignment with this program will be #2. The program will exit after you enter your response (you can enter a blank response or any other response you wish). Then click on Quit Program (upper right-hand corner).
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RESPONSE -->
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ᙬСLzkՍxz Student Name: assignment #002
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21:08:00 `q001. Note that there are 14 questions in this assignment.
If an object moves 12 meters in 4 seconds, then at what average rate is the object moving? Explain how you obtained your result in terms of commonsense images.......!!!!!!!!...................................
RESPONSE --> 12 / 4 = 3m/sec. If the total distance moved is 12m, and it took 4 seconds to get there, then the object must move an average of 3m per second.
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21:08:28 Moving 12 meters in 4 seconds, we move an average of 3 meters every second. We can imagine dividing up the 12 meters into four equal parts, one for each second. Each part will have 3 meters, corresponding to the distance moved in 1 second, on the average.
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RESPONSE --> ok mine wasn't so visualized, but still comprehendable.
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21:09:49 `q002. How is this problem related to the concept of a rate?
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RESPONSE --> 3m per sec is the rate of the object. Dividing the distance by the time gives the rate at which the item moved.
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21:09:57 A rate is obtained by dividing the change in a quantity by the change in another quantity on which is dependent. In this case we divided the change in position by the time during which that change occurred.
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RESPONSE --> ok
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21:11:46 `q003. Is object position dependent on time or is time dependent on object position?
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RESPONSE --> I think the time is dependent on object position. The distance traveled directly affects the time it takes to get it there. Since the amount of time has no influence on the position, the position of the object is the independent variable.
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21:13:20 Object position is dependent on time--the clock runs whether the object is moving or not so time is independent of position. Clock time is pretty much independent of anything else.
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RESPONSE --> Oops I'm completely backwards. I thought of the time being independent, but what I should have done was picture how I would graph the two. Since the time would be on the x axis, it would be the independent variable.
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21:14:01 `q004. So the rate here is the average rate at which position is changing with respect to clock time. Explain what concepts, if any, you missed in your explanations.
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RESPONSE --> The only concepts I feel I missed was the confusion on the independent and dependent variables. Other than that I don't feel I misunderstood anything.
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21:14:14 You should always self-critique your work in this manner. Always critique your solutions by describing any insights you had or errors you makde, and by explaining how you can make use of the insight or how you now know how to avoid certain errors. Also pose for the instructor any question or questions that you have related to the problem or series of problems.
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RESPONSE --> ok
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21:18:02 `q005. If an object is displaced -6 meters in three seconds, then what is the average speed of the object what is its average velocity? Explain how you obtained your result in terms of commonsense images and ideas.
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RESPONSE --> -6m / 3sec = -2m/s is the average speed of the object. I'm not sure if the question is asking for both the speed and the velocity... As before, if the object moved a total of -6m, then dividing it by the time gives the number of meters per second that the object was displaced.
Though I'm not sure, since the distance and time given are not intervals, then the velocity of the object should also be -6 / 3 = -2m/s..................................................
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21:19:43 Speed is the average rate at which distance changes, and distance cannot be negative. Therefore speed cannot be negative. Velocity is the average rate at which position changes, and position changes can be positive or negative.
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RESPONSE --> Since the distance was negative, there was actually no need for any calculations. But since velocity can be either positive or negative, I don't see why there isn't any calculations done for the question of the velocity of the object.
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21:21:33 `q006. If `ds stands for the change in the position of an object and `dt for the time interval during which the position changes, then what expression stands for the average velocity vAve of the object during this time interval?
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RESPONSE --> 'ds / 'dt is the equation used for the average velocity.
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21:21:41 Average velocity is rate of change of position. Change in position is `ds and change in clock tim is `dt, so vAve = `ds / `dt.
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RESPONSE --> ok
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21:22:42 `q007. How do you write the expressions `ds and `dt on your paper?
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RESPONSE --> Since 'ds and 'dt stand for the change in an object's distance and time, you can right the expressions with a delta symbol as opposed to the 'ds and 'dt.
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21:22:58 You use the Greek capital Delta symbol Delta. `d is often used here because the symbol for Delta is not interpreted correctly by some Internet browsers. You should get in the habit of thinking and writing Delta when you see `d. You may use either `d or Delta when submitting work and answering questions.
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RESPONSE --> ok
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21:24:16 `q008. If an object changes position at an average rate of 5 meters/second for 10 seconds, then how far does it move?
How is this problem related to the concept of a rate?......!!!!!!!!...................................
RESPONSE --> 5m/s * 10s = 50m/s. This problem relates to the concept of a rate because it gives the rate at which the object moves (5m/s) and the total amount of time it is moving.
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21:26:51 In this problem you are given the rate at which position changes with respect to time, and you are given the time interval during which to calculate the change in position. Given the rate at which one quantity changes with respect to another, and the change in the second quantity, how do we obtain the resulting change in the first?
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RESPONSE --> I've read this question 3 times and am still not sure as to what exactly it is asking...
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21:28:46 `q009. If vAve stands for the rate at which the position of the object changes (also called velocity) and `dt for the time interval during which the change in position is to be calculated, then how to we write the expression for the change `ds in the position?
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RESPONSE --> vAve * 'dt = 'ds. I'm guessing on this. I just took the equation for velocity ('ds / 'dt) and rearranged it to solve for 'ds.
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21:29:21 To find the change in a quantity we multiply the rate by the time interval during which the change occurs. We therefore obtain the change in position by multiplying the velocity by the time interval: `ds = vAve * `dt. The units of this calculation pretty much tell us what to do: Just as when we multiply pay rate by time (dollar / hr * hours of work) or automobile velocity by the time interval (miles / hour * hour), when we multiply vAve, in cm / sec or meters / sec or whatever, by `dt in seconds, we get displacement in cm or meters, or whatever, depending on the units of distance used.
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RESPONSE --> ok
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21:32:07 `q010. Explain how the quantities average velocity vAve, time interval `dt and displacement `ds are related by the definition of a rate, and how this relationship can be used to solve the current problem problem.
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RESPONSE --> The rate is found by dividing the change in position 'ds by the change in time 'dt, which gives the average velocity of the object. By dividing the dependent variable by the independent one, we get the rate of the object.
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21:32:16 vAve is the average rate at which position changes. The change in position is the displacement `ds, the change in clock time is `dt, so vAve = `ds / `dt.
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RESPONSE --> ok
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21:33:22 `q011. The basic rate relationship vAve = `ds / `dt expresses the definition of average velocity vAve as the rate at which position s changes with respect to clock time t. What algebraic steps do we use to solve this equation for `ds, and what is our result?
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RESPONSE --> vAve = `ds / `dt To solve for 'ds, we need to isolate it, which means that we'd have to multiply 'dt by vAve, resulting in 'ds = vAve * 'dt.
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21:33:32 To solve vAve = `ds / `dt for `ds, we multiply both sides by `dt. The steps:
vAve = `ds / `dt. Multiply both sides by `dt: vAve * `dt = `ds / `dt * `dt Since `dt / `dt = 1 vAve * `dt = `ds . Switching sides we have `ds = vAve * `dt.......!!!!!!!!...................................
RESPONSE --> ok
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21:40:07 `q012. How is this result related to our intuition about the meanings of the terms average velocity, displacement and clock time?
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RESPONSE --> I'm not entirely sure as to what the question's asking. The only thing I could think of is that the three variables play off each other and each can be found by using the initial vAve = 'ds / 'dt equation.
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21:41:19 Our most direct intuition about velocity probably comes from watching an automobile speedometer. We know that if we multiply our average velocity in mph by the duration `dt of the time interval during which we travel, we get the distance traveled in miles. From this we easily extend the idea. Whenever we multiply our average velocity by the duration of the time interval, we expect to obtain the displacement, or change in position, during that time interval.
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RESPONSE --> Oh ok I see what the question was pertaining to now. I couldn't figure out what it meant in order to explain how to get the displacement or the change in position.
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21:43:47 `q013. What algebraic steps do we use to solve the equation vAve = `ds / `dt for `dt, and what is our result?
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RESPONSE --> As before, we need to get 'dt by itself. I'm sure my process is kind of excessive and it can be done easier, but this is the first thing that came to my mind. vAve = 'ds/'dt multiply 'dt by both sides to get 'dt * vAve = 'ds. Then to get 'dt by itself, divide vAve on both sides to get 'dt = 'ds / vAve.
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21:44:30 To solve vAve = `ds / `dt for `dt, we must get `dt out of the denominator. Thus we first multiply both sides by the denominator `dt. Then we can see where we are and takes the appropriate next that. The steps:
vAve = `ds / `dt. Multiply both sides by `dt: vAve * `dt = `ds / `dt * `dt Since `dt / `dt = 1 vAve * `dt = `ds. We can now divide both sides by vAve to get `dt = `ds / vAve.......!!!!!!!!...................................
RESPONSE --> I'm glad what I thought was excessive was actually right. For some reason I was thinking there was an easier way to solving this.
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21:46:50 `q014. How is this result related to our intuition about the meanings of the terms average velocity, displacement and clock time?
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RESPONSE --> If we divide our distance traveled by the average velocity, then we should be able to obtain the time interval that it took the object to travel at that rate or speed.
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21:47:10 If we want to know how long it will take to make a trip at a certain speed, we know to divide the distance in miles by the speed in mph. If we divide the number of miles we need to travel by the number of miles we travel in hour, we get the number of hours required. We extend this to the general concept of dividing the displacement by the velocity to get the duration of the time interval.
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RESPONSE --> ok
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