course Phy 201 I have not received answers to the questions I have sent through email over the past week. Am I supposed to submit questions in a different format?Fridays and Saturdays are the days I set aside to work on Physics, but the server was down Friday evening and that is why I am late submitting some of these assignments. `ńKQw_
......!!!!!!!!...................................
12:49:18 `q001 Note that there are 10 questions in this assignment. At a certain instant the speedometer of a car reads 5 meters / second (of course cars in this country generally read speeds in miles per hour and km per hour, not meters / sec; but they could easily have their faces re-painted to read in meters/second, and we assume that this speedometer has been similarly altered). Exactly 4 seconds later the speedometer reads 25 meters/second (that, incidentally, indicates very good acceleration, as you will understand later). At what average rate is the speed of the car changing with respect to clock time?
......!!!!!!!!...................................
RESPONSE --> Average rate of speed = `dv / `dt =(25m/s - 5 m/s) / 4s =5m/s/s confidence assessment: 2
.................................................
......!!!!!!!!...................................
12:49:31 The rate of change of the speed with respect clock time is equal to the change in the speed divided by the change in the clock time. So we must ask, what is the change in the speed, what is the change in the clock time and what therefore is the rate at which the speed is changing with respect to clock time? The change in speed from 5 meters/second to 25 meters/second is 20 meters/second. This occurs in a time interval lasting 4 seconds. The average rate of change of the speed is therefore (20 meters/second)/(4 seconds) = 5 meters / second / second. This means that on the average, per second, the speed changes by 5 meters/second.
......!!!!!!!!...................................
RESPONSE --> ok self critique assessment: 3
.................................................
......!!!!!!!!...................................
12:54:54 `q002. Explain in commonsense terms of the significance for an automobile of the rate at which its velocity changes. Do you think that a car with a more powerful engine would be capable of a greater rate of velocity change?
......!!!!!!!!...................................
RESPONSE --> velocity changes as the speedometer needle moves up and down. If the needle moves from 50mph to 60 mph the velocity is increasing because the distance covered in one hour increases. This increase could be constant, increasing or decreasing depending the time it takes for the needle to move from one position to another. A car with a more powerful engine would be capapble of a greater rate velocity change because it would be able to accelerate faster and the time it would take for the needle to move from one poipnt to another would take less time. confidence assessment: 2
.................................................
......!!!!!!!!...................................
12:55:11 A car whose velocity changes more rapidly will attain a given speed in a shorter time, and will be able to 'pull away from' a car which is capable of only a lesser rate of change in velocity. A more powerful engine, all other factors (e.g., weight and gearing) being equal, would be capable of a greater change in velocity in a given time interval.
......!!!!!!!!...................................
RESPONSE --> ok self critique assessment: 3
.................................................
......!!!!!!!!...................................
12:56:41 `q003. Explain how we obtain the units meters / second / second in our calculation of the rate of change of the car's speed.
......!!!!!!!!...................................
RESPONSE --> rate of change = `dv / `dt So the rate of change = (m/s) / (s) These units do not cancel each other out. confidence assessment: 2
.................................................
......!!!!!!!!...................................
12:56:48 When we divide the change in velocity, expressed in meters/second, by the duration of the time interval in seconds, we get units of (meters / second) / second, often written meters / second / second.
......!!!!!!!!...................................
RESPONSE --> ok self critique assessment: 3
.................................................
......!!!!!!!!...................................
12:58:17 `q004. The unit (meters / second) / second is actually a complex fraction, having a numerator which is itself a fraction. Such a fraction can be simplified by multiplying the numerator by the reciprocal of the denominator. We thus get (meters / second) * (1/ second). What do we get when we multiply these two fractions?
......!!!!!!!!...................................
RESPONSE --> m/s/s confidence assessment: 2
.................................................
......!!!!!!!!...................................
12:59:16 Multiplying the numerators we get meters * 1; multiplying the denominators we get second * second, which is second^2. Our result is therefore meters * 1 / second^2, or just meters / second^2. If appropriate you may at this point comment on your understanding of the units of the rate of change of velocity.
......!!!!!!!!...................................
RESPONSE --> It's hard for me to understand this without using actual numbers. But it is the same thing as m/s/s, correct? self critique assessment: 2
.................................................
......!!!!!!!!...................................
13:02:53 `q004. If the velocity of an object changes from 10 m/s to -5 m/s during a time interval of 5 seconds, then at what average rate is the velocity changing?
......!!!!!!!!...................................
RESPONSE --> average rate of change of v = `dv /`dt =(10m/s - -5m/s) / 5s =3m/s/s confidence assessment: 1
.................................................
......!!!!!!!!...................................
13:04:19 We see that the velocity changes from 10 meters/second to -5 meters/second, a change of -15 meters / second, during a five-second time interval. A change of -15 m/s during a 5 s time interval implies an average rate of -15 m/s / (5 s) = -3 (m/s)/ s = -3 m/s^2. This is the same as (-3 m/s) / s, as we saw above. So the velocity is changing by -3 m/s every second.
......!!!!!!!!...................................
RESPONSE --> I should have written it as (-5 -10) rather than (10- -5). This shows that the velocity is decreasing. self critique assessment: 2
.................................................
......!!!!!!!!...................................
13:05:11 `q005. You should have noted that velocity can be positive or negative, as can the change in velocity or the rate at which velocity changes. The average rate at which a quantity changes with respect to time over a given time interval is equal to the change in the quantity divided by the duration of the time interval. In this case we are calculating the average rate at which the velocity changes. If v represents velocity then we we use `dv to represent the change in velocity and `dt to represent the duration of the time interval. What expression do we therefore use to express the average rate at which the velocity changes?
......!!!!!!!!...................................
RESPONSE --> `dvAve = `dv/`dt confidence assessment: 2
.................................................
......!!!!!!!!...................................
13:06:06 The average rate would be expressed by [ave rate of velocity change with respect to clock time] = `dv / `dt. The expression [ave rate of velocity change with respect to clock time] is pretty cumbersome so we give it a name. The name we give it is 'average acceleration', abbreviated by the letter aAve. Using a to represent acceleration, write down the definition of average acceleration. The definition of average acceleration is aAve = `dv / `dt. Please make any comments you feel appropriate about your understanding of the process so far.
......!!!!!!!!...................................
RESPONSE --> ok, thats easier to write. I was trying to abbreviate it as `dvAve meaning a change in the average rate of velocity self critique assessment: 2
.................................................
......!!!!!!!!...................................
13:19:31 `q006. If a runner is moving at 6 meters / sec at clock time t = 1.5 sec after starting a race, and at 9 meters / sec at clock time t = 3.5 sec after starting, then what is the average acceleration of the runner between these two clock times?
......!!!!!!!!...................................
RESPONSE --> aAve = `dv / `dt = (9m/s - 6 m/s) / (3.5s - 1.5 s) = (3m/s) / (2.0 m/s) = 1.5 m/s/s confidence assessment: 2
.................................................
......!!!!!!!!...................................
13:21:31 `q006b. What therefore is the average rate at which the velocity is changing during this time interval?
......!!!!!!!!...................................
RESPONSE --> The same thing: 1.5 m/s/s confidence assessment: 1
.................................................
......!!!!!!!!...................................
13:21:58 We see that the runner's velocity changes from 6 meters/second to 9 meters/second, a change of `dv = 9 m/s - 6 m/s = 3 m/s, during a time interval their runs from t = 1.5 sec to t = 3.5 sec so that the duration of the interval is `dt = 3.5 s - 1.5 s = 2.0 s. The rate at which the velocity changes is therefore 3 m/s / (2.0 s) = 1.5 m/s^2. Please comment if you wish on your understanding of the problem at this point.
......!!!!!!!!...................................
RESPONSE --> ok self critique assessment: 3
.................................................
......!!!!!!!!...................................
13:46:58 `q007. On a graph of velocity vs. clock time, we can represent the two events of this problem by two points on the graph. The first point will be (1.5 sec, 6 meters/second) and the second point will be (3.5 sec, 9 meters / sec). What is the run between these points and what does it represent? What is the rise between these points what does it represent? What is the slope between these points what does it represent?
......!!!!!!!!...................................
RESPONSE --> rise: 3 Rise represents the change in velocity slope = rise/run = 3/2 = 1.5 Slope represents the average acceleration confidence assessment: 2
.................................................
......!!!!!!!!...................................
13:47:17 The rise from the first point to the second is from 6 meters/second to 9 meters/second, or 3 m/s. This represents the change `dv in velocity. The run is from 1.5 seconds to 3.5 seconds, or 2 seconds, and represents the change `dt in clock time. The slope, being the rise divided by the run, is 3 m/s / (2 sec) = 1.5 m/s^2. This slope represents `dv / `dt, which is the average acceleration during the time interval. You may if you wish comment on your understanding to this point.
......!!!!!!!!...................................
RESPONSE --> ok self critique assessment: 3
.................................................
......!!!!!!!!...................................
13:48:37 `q008. In what sense does the slope of any graph of velocity vs. clock time represent the acceleration of the object? For example, why does a greater slope imply greater acceleration?
......!!!!!!!!...................................
RESPONSE --> The slope will equal the average accelerations, so if there is a greater slope, there will be a greater acceleration. confidence assessment: 2
.................................................
......!!!!!!!!...................................
13:48:48 Since the rise between two points on a graph of velocity vs. clock time represents the change in `dv velocity, and since the run represents the change `dt clock time, the slope represents rise / run, or change in velocity /change in clock time, or `dv / `dt. This is the definition of average acceleration.
......!!!!!!!!...................................
RESPONSE --> ok self critique assessment: 3
.................................................
......!!!!!!!!...................................
13:53:04 `q009. This is the same situation as in the preceding problem: An automobile coasts down a hill with a constant slope. At first its velocity increases at a very nearly constant rate. After it attains a certain velocity, air resistance becomes significant and the rate at which velocity changes decreases, though the velocity continues to increase. Describe a graph of velocity vs. clock time for this automobile (e.g., neither increasing nor decreasing; increasing at an increasing rate, a constant rate, a decreasing rate; decreasing at an increasing, constant or decreasing rate; the description could be different for different parts of the graph).
......!!!!!!!!...................................
RESPONSE --> The first part of the graph is increasing at a nearly constant rate. Then as the air restistance becomes significant, the graph is increasing at a decreasing rate. confidence assessment: 2
.................................................
......!!!!!!!!...................................
13:53:30 Your graph should have velocity as the vertical axis and clock time as the horizontal axis. The graph should be increasing since the velocity starts at zero and increases. At first the graph should be increasing at a constant rate, because the velocity is increasing at a constant rate. The graph should continue increasing by after a time it should begin increasing at a decreasing rate, since the rate at which the velocity changes begins decreasing due to air resistance. However the graph should never decrease, although as air resistance gets greater and greater the graph might come closer and closer to leveling off. Critique your solution by describing or insights you had or insights you had and by explaining how you now know how to avoid those errors.
......!!!!!!!!...................................
RESPONSE --> ok self critique assessment: 3
.................................................
......!!!!!!!!...................................
13:55:21 `q010. An automobile coasts down a hill with a constant slope. At first its velocity increases at a very nearly constant rate. After it attains a certain velocity, air resistance becomes significant and the rate at which velocity changes decreases, though the velocity continues to increase. Describe a graph of acceleration vs. clock time for this automobile (e.g., neither increasing nor decreasing; increasing at an increasing rate, a constant rate, a decreasing rate; decreasing at an increasing, constant or decreasing rate; the description could be different for different parts of the graph).
......!!!!!!!!...................................
RESPONSE --> The first part of the graph is increasing at a nearly constant rate. Then as the air restistance becomes significant, the graph is increasing at a decreasing rate. Isn't this the same question as #9? confidence assessment: 2
.................................................
......!!!!!!!!...................................
13:59:08 Your graph should have acceleration as the vertical axis and clock time as the horizontal axis. At first the graph should be neither increasing nor decreasing, since it first the acceleration is constant. Then after a time the graph should begin decreasing, which indicates the decreasing rate at which velocity changes as air resistance begins having an effect. An accurate description of whether the graph decreases at a constant, increasing or decreasing rate is not required at this point, because the reasoning is somewhat complex and requires knowledge you are not expected to possess at this point. However it is noted that the graph will at first decrease more and more rapidly, and then less and less rapidly as it approaches the t axis. In answer to the following question posed at this point by a student: Can you clarify some more the differences in acceleration and velocity? ** Velocity is the rate at which position changes and the standard units are cm/sec or m/sec. Acceleration is the rate at which velocity changes and its standard units are cm/s^2 or m/s^2. Velocity is the slope of a position vs. clock time graph. Acceleration is the slope of a velocity vs. clock time graph. **
......!!!!!!!!...................................
RESPONSE --> This answer is different from that of #9, but the questions were the same. Why would the graph ""at first decrease more and more rapidly, and then less and less rapidly as it approaches the t axis"" ? self critique assessment: 2
.................................................
"