course PHY 121 I am only copying part of this file to send to you because I have already sent the file once today.As to the question you posed for experiment three - I am pretty sure the points were 20.5, 37 and 16.4, 33. To be quite honest I am not sure if this is from the right graph. Let me know what you think
.................................................
詨簿ぉ桕蔞中跐服枖见儕见岁 Student Name: assignment #002
.................................................
......!!!!!!!!...................................
15:21:50 `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 --> 12m / 4s = 3m/s I visualized and drew a 12 m ramp with a toy car moving at a rate of 12meters per 4 secs, then broke down into secs would be 3 meters per sec.
.................................................
......!!!!!!!!...................................
15:21:58 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.
......!!!!!!!!...................................
RESPONSE --> correct
.................................................
......!!!!!!!!...................................
16:29:29 `q002. How is this problem related to the concept of a rate?
......!!!!!!!!...................................
RESPONSE --> This is definately a simple rate, being easy and natural. Most people of average intelligence could respond correctly without writing it out.
.................................................
......!!!!!!!!...................................
16:31:09 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.
......!!!!!!!!...................................
RESPONSE --> The problem was so simple that I did not realize that it fell into the average rate definition.
.................................................
......!!!!!!!!...................................
16:37:49 `q003. Is object position dependent on time or is time dependent on object position?
......!!!!!!!!...................................
RESPONSE --> Niether is dependent on either. However, both are dependent on velocity.
.................................................
......!!!!!!!!...................................
16:39:04 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.
......!!!!!!!!...................................
RESPONSE --> I must have not interpreted the synopsis notes correctly.
.................................................
......!!!!!!!!...................................
16:42:19 `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.
......!!!!!!!!...................................
RESPONSE --> Average Rate = postion change v. clock time
.................................................
......!!!!!!!!...................................
16:42:50 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.
......!!!!!!!!...................................
RESPONSE --> ok
.................................................
......!!!!!!!!...................................
16:47:42 `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.
......!!!!!!!!...................................
RESPONSE --> 6 Meters / 3 seconds = 2 M/S. As with the previous problem, I visualized a 6 meters of ramp with an object moving the full length in 3 seconds. There is no mention here of velocity or acceleration. Therefore, one can only conclude that the velocity was at a constant rate. Each second the object moved 2 meters to arrive at the end in 6 seconds.
.................................................
......!!!!!!!!...................................
16:48:15 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.
......!!!!!!!!...................................
RESPONSE --> okay
.................................................
......!!!!!!!!...................................
01:18:22 `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?
......!!!!!!!!...................................
RESPONSE --> `ds/`dt = vAve
.................................................
......!!!!!!!!...................................
01:18:34 Average velocity is rate of change of position. Change in position is `ds and change in clock tim is `dt, so vAve = `ds / `dt.
......!!!!!!!!...................................
RESPONSE --> ok
.................................................
......!!!!!!!!...................................
01:20:02 `q007. How do you write the expressions `ds and `dt on your paper?
......!!!!!!!!...................................
RESPONSE --> Using a triangle (i think it is called delta), instead of `d this means change.
.................................................
......!!!!!!!!...................................
01:20:34 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.
......!!!!!!!!...................................
RESPONSE --> ok
.................................................
......!!!!!!!!...................................
01:23:33 `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 = 50meters, the seconds would cancel. The problem gives us the m/s and we are asked to lets say work backwards and multiply to get our answer.
.................................................
......!!!!!!!!...................................
01:24:59 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?
......!!!!!!!!...................................
RESPONSE --> by multiplying the change in the first by the time interval.
.................................................
......!!!!!!!!...................................
01:26:47 `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?
......!!!!!!!!...................................
RESPONSE --> `ds = vAve * `dt
.................................................
......!!!!!!!!...................................
01:26:54 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.
......!!!!!!!!...................................
RESPONSE --> ok
.................................................
......!!!!!!!!...................................
01:30:56 `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.
......!!!!!!!!...................................
RESPONSE --> We are using the change in position and time to obtain the vAve by division.
.................................................
......!!!!!!!!...................................
01:31:03 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.
......!!!!!!!!...................................
RESPONSE --> ok
.................................................
......!!!!!!!!...................................
01:32:32 `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?
......!!!!!!!!...................................
RESPONSE --> We multiply the avg. Velocity by the change in time which will give us the change in postion, The units of time will cancel and we will be left with the units of position.
.................................................
......!!!!!!!!...................................
01:32:47 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 --> correct
.................................................
......!!!!!!!!...................................
01:41:25 `q012. How is this result related to our intuition about the meanings of the terms average velocity, displacement and clock time?
......!!!!!!!!...................................
RESPONSE --> b/c the avg velocity if the avg. rate at which the position of an object changes between two clock times.
.................................................
......!!!!!!!!...................................
01:43:12 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.
......!!!!!!!!...................................
RESPONSE --> The speedometer is an excellent commonsense mechanism to apply to these concepts.
.................................................
......!!!!!!!!...................................
01:45:53 `q013. What algebraic steps do we use to solve the equation vAve = `ds / `dt for `dt, and what is our result?
......!!!!!!!!...................................
RESPONSE --> vAve = `ds / `dt vAve * `ds = `ds * `ds / `dt The change in position would cancel each other out on the right side of the euation and we would be left with: `dt = vAve * `ds
.................................................
......!!!!!!!!...................................
01:47:42 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 missed this because I forgot to get the change in time out of the denominator.
.................................................
......!!!!!!!!...................................
01:49:16 `q014. How is this result related to our intuition about the meanings of the terms average velocity, displacement and clock time?
......!!!!!!!!...................................
RESPONSE --> We know that if we divide our distqance traveled by the mph, we will get the time it takes us to get there
.................................................
......!!!!!!!!...................................
01:49:28 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.
......!!!!!!!!...................................
RESPONSE --> ok
.................................................