assignment 2

Your work has been received. Please scroll to the end of the document to see any inserted notes (in boldface) and a note at the end. The note at the end of the file will confirm that the file has been reviewed; be sure to read that note. If there is no note at the end, notify the instructor through the Submit Work form, and include the date of the posting to your access page.

From the understanding you have expressed so far, I think the following might help you clarify the situation:

The complete phrase is always 'average rate of change of ___ with respect to ____'. The meaning depends on what words are used to fill in the blanks.

'average rate of change of position with respect to clock time' is average velocity. 'average rate of change of velocity with respect to clock time' is average acceleration.

There are a number of other rates involved in physics.

The average rate of change of A with respect to B is ave rate = change in A / change in B.

Therefore

average velocity = ave rate of change of position with respect to clock time = change in position / change in clock time

and

average acceleration = ave rate of change of velocity with respect to clock time = change in velocity / change in clock time.

ÇïÜñ¡ßýòÙÿäÕ‹Âúš´~wStudent assignment #002

.................................................

......!!!!!!!!...................................

15:14:36 `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 --> since it's initial state is 0 cm at 0 seconds, change is simpily 12cm/4sec. 3cm/sec is therefore the average rate.

.................................................

......!!!!!!!!...................................

15:15:43 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 --> check, although I'm not sure why I though cm was the unit.

.................................................

......!!!!!!!!...................................

15:16:51 `q002. How is this problem related to the concept of a rate?

......!!!!!!!!...................................

RESPONSE --> a rate is a quantity vs time graph. As time continues the quantity can increase, decrease, or stay the same.

.................................................

......!!!!!!!!...................................

15:17:16 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 --> check

.................................................

......!!!!!!!!...................................

15:19:12 `q003. Is object position dependent on time or is time dependent on object position?

......!!!!!!!!...................................

RESPONSE --> object position is dependent on time. An example of this is change in position, but to further explain. If time was dependent on the object, that would indicate that if the object didn't move, then time didn't pass. That doesn't happen.

This is an excellent statement, which I'm going to post for everyone. The post won't include your name, but if you want to include your name or at least your initials (your choice) email me and I'll be glad to add it.

.................................................

......!!!!!!!!...................................

15:19:33 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 --> check

.................................................

......!!!!!!!!...................................

15:21:27 `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 --> I pretty much got the change in position vs time.

.................................................

......!!!!!!!!...................................

15:23:38 `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, that going in reverse however, velocity isn't affected by this, but speed is. 6/3 = 2 so velocity is 2m/sec but speed is -2m/sec.

.................................................

......!!!!!!!!...................................

15:31:14 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 --> ah, velocity can be negetive, it is speed that is absolute. Okay, because distance is always positive, speed must be too. Change in position can be negative if the object moves back.

.................................................

......!!!!!!!!...................................

15:32:31 `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 change in pos/change in time

.................................................

......!!!!!!!!...................................

15:32:47 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 --> check

.................................................

......!!!!!!!!...................................

15:34:19 `q007. How do you write the expressions `ds and `dt on your paper?

......!!!!!!!!...................................

RESPONSE --> deltaS/deltaT delta is a greek symbol that resembles a triangle.

.................................................

......!!!!!!!!...................................

15:34:42 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 --> check

.................................................

......!!!!!!!!...................................

15:36:03 `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 --> 5meters/second x 10seconds. First, you can think of the seconds unit cancelling each other. 50 meters.

.................................................

......!!!!!!!!...................................

15:38:41 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 --> check

.................................................

......!!!!!!!!...................................

15:44:09 `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 --> vAve x 'dt = 'ds

.................................................

......!!!!!!!!...................................

15:44:25 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 --> check

.................................................

......!!!!!!!!...................................

15:46:09 `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 --> velocity or vAve can be solved by 'ds/'dt velocity over/'dt can give you an average rate, or that is to my understanding of this concept.

Close, but it should be

ave rate of change of velocity with respect to clock time = change in vel / change in clock time.

.................................................

......!!!!!!!!...................................

15:48:27 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 --> so, average rate is simply vAve.

.................................................

......!!!!!!!!...................................

15:49:37 `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 --> vAve = 'ds/'dt vAve('dt) = 'ds

.................................................

......!!!!!!!!...................................

15:49:47 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 --> check

.................................................

......!!!!!!!!...................................

15:51:49 `q012. How is this result related to our intuition about the meanings of the terms average velocity, displacement and clock time?

......!!!!!!!!...................................

RESPONSE --> displacement or the change in position is divided by the change in clock time. This will result in the average velocity. It can be thought of as how long did it take to go from here to there, which is our velocity.

.................................................

......!!!!!!!!...................................

15:52:18 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 --> check

.................................................

......!!!!!!!!...................................

15:53:52 `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 x 'dt = 'ds 'dt = 'ds/vAve

.................................................

......!!!!!!!!...................................

15:54:05 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 --> check

.................................................

......!!!!!!!!...................................

15:56:31 `q014. How is this result related to our intuition about the meanings of the terms average velocity, displacement and clock time?

......!!!!!!!!...................................

RESPONSE --> we think of it taking so long to go so many miles. Thus, we take how much we have gone and divide it by how far we should be in one hour. This tells us how long we have been driving.

.................................................

......!!!!!!!!...................................

15:57:08 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 --> check, perhaps.

.................................................

"

You did really well on these questions.

We'll be sorting out the rate concept over the next couple of assignments, but so far you've got an excellent handle on it. Hopefully my notes will further clarify the situation.