course Phy 201 ýO¶¯š¡}ÛÔrðñî¥x‡|~¡Åassignment #002
......!!!!!!!!...................................
18:56:09 `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 --> average rate = 12m/4s average rate =3m/s To determine the rate of an object that moves 12 meters in 4 seconds then divide the chande in distance (12 meters) by the change in time (4 seconds). Since there are no canceling unit the we figure that 12 divided by 4 equals 3 and that the meters are still divided by the second thus we get the result of 3 m/s confidence assessment: 3
.................................................
......!!!!!!!!...................................
18:56:41 06-25-2008 18:56:41 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.
......!!!!!!!!...................................
NOTES ------->
.......................................................!!!!!!!!...................................
18:57:51 `q002. How is this problem related to the concept of a rate?
......!!!!!!!!...................................
RESPONSE --> This problem is related in rate because it was a measurement of the change in something divided by the change in something else. In this case it was the change in position divided by the change in time. confidence assessment: 3
.................................................
......!!!!!!!!...................................
18:58:04 06-25-2008 18:58:04 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.
......!!!!!!!!...................................
NOTES ------->
.......................................................!!!!!!!!...................................
19:00:20 `q003. Is object position dependent on time or is time dependent on object position?
......!!!!!!!!...................................
RESPONSE --> Object position is dependent on time. confidence assessment: 3
.................................................
......!!!!!!!!...................................
19:00:35 06-25-2008 19:00:35 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.
......!!!!!!!!...................................
NOTES ------->
.......................................................!!!!!!!!...................................
19:01:50 `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 --> Although I knew that position was dependent on time, the explanation helped me to better understand why in the fact that time is always changing. confidence assessment: 3
.................................................
......!!!!!!!!...................................
19:01:58 06-25-2008 19:01:58 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.
......!!!!!!!!...................................
NOTES ------->
.......................................................!!!!!!!!...................................
19:06:01 `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 --> Average velocity = 6 meters / 3 seconds Average velocity = 2 m/s To determine the rate of an object that moves -6 meters in 3 seconds then divide the absolute value of change in distance (-6 meters) by the change in time (3 seconds). Since there are no canceling unit the we figure that 6 divided by 3 equals 2 and that the meters are still divided by the second thus we get the result of 2 m/s confidence assessment: 3
.................................................
......!!!!!!!!...................................
19:06:17 06-25-2008 19:06:17 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.
......!!!!!!!!...................................
NOTES ------->
.......................................................!!!!!!!!...................................
19:07:30 `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 --> vAve = 'ds/'dt confidence assessment: 3
.................................................
......!!!!!!!!...................................
19:07:41 06-25-2008 19:07:41 Average velocity is rate of change of position. Change in position is `ds and change in clock tim is `dt, so vAve = `ds / `dt.
......!!!!!!!!...................................
NOTES ------->
.......................................................!!!!!!!!...................................
19:09:56 `q007. How do you write the expressions `ds and `dt on your paper?
......!!!!!!!!...................................
RESPONSE --> Using the greek letter Delta followed by a s above the divide line. Under the divide line is letter Delta followed by a t. confidence assessment: 3
.................................................
......!!!!!!!!...................................
19:10:20 06-25-2008 19:10:20 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.
......!!!!!!!!...................................
NOTES ------->
.......................................................!!!!!!!!...................................
19:12:55 `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 = 'ds/10s 10s * 5m/s = 'ds The seconds units cancel out 10 * 5m = 'ds 50m = 'ds This is related to the concept of rate in which the value of rate had been given along with the change in time which led to the determining the change in position. confidence assessment: 3
.................................................
......!!!!!!!!...................................
19:14:00 06-25-2008 19:14:00 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?
......!!!!!!!!...................................
NOTES ------->
.......................................................!!!!!!!!...................................
19:14:35 `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 confidence assessment: 3
.................................................
......!!!!!!!!...................................
19:14:48 06-25-2008 19:14:48 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.
......!!!!!!!!...................................
NOTES ------->
.......................................................!!!!!!!!...................................
19:16:58 `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 --> vAve is considered to be the rate which is deifined as a change in a quantity ('ds) divided by the change in another quantity ('dt). This relationship is used to determine the velocity of an object that change position over a given period of time. confidence assessment: 3
.................................................
......!!!!!!!!...................................
19:17:10 06-25-2008 19:17:10 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.
......!!!!!!!!...................................
NOTES ------->
.......................................................!!!!!!!!...................................
19:20:05 `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 Following order of operations in solving for 'ds, we would move the 'dt to the other side of the equals sign in which the operation it was supposed to perform changes to its opposite resulting in having 'dt be a multiplier. 'dt * vAve = 'ds Multiply vAve by 'dt and the result is equal to 'ds. confidence assessment: 3
.................................................
......!!!!!!!!...................................
19:20:19 06-25-2008 19:20:19 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.
......!!!!!!!!...................................
NOTES ------->
.......................................................!!!!!!!!...................................
19:21:23 `q012. How is this result related to our intuition about the meanings of the terms average velocity, displacement and clock time?
......!!!!!!!!...................................
RESPONSE --> vAve is average velocity. 'ds is displacement 'dt is the time interval. confidence assessment: 3
.................................................
......!!!!!!!!...................................
19:21:35 06-25-2008 19:21:35 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.
......!!!!!!!!...................................
NOTES ------->
.......................................................!!!!!!!!...................................
19:22:51 `q013. What algebraic steps do we use to solve the equation vAve = `ds / `dt for `dt, and what is our result?
......!!!!!!!!...................................
RESPONSE --> Multiply both sides by 'dt and then divide both sides by vAve resulting in 'dt = 'ds/vAve. confidence assessment: 3
.................................................
......!!!!!!!!...................................
19:23:05 06-25-2008 19:23:05 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.
......!!!!!!!!...................................
NOTES ------->
.......................................................!!!!!!!!...................................
19:24:34 `q014. How is this result related to our intuition about the meanings of the terms average velocity, displacement and clock time?
......!!!!!!!!...................................
RESPONSE --> distance that is dependent on speed results in a time interval. A distance traveled at a certain speed results in time spent. confidence assessment: 3
.................................................
......!!!!!!!!...................................
19:24:44 06-25-2008 19:24:44 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.
......!!!!!!!!...................................
NOTES ------->
.................................................
course Phy 201 Would like to have feedback. Thanks, Joshua assignment #002002. `Query 2
......!!!!!!!!...................................
20:29:05 How is acceleration an example of a rate of change?
......!!!!!!!!...................................
RESPONSE --> Acceleration is an example of rate change because it is defined as the change in velocity divided by the time interval. Rate is a change in a quantity divided by the change in another quantity. confidence assessment: 3
.................................................
......!!!!!!!!...................................
20:29:22 06-25-2008 20:29:22 ** Velocity is the rate of change of position. Acceleration is rate of change of velocity--change in velocity divided by the time period. To find the acceleration from a v vs. t graph you take the rise, which represents the change in the average velocity, and divide by the run, which represents the change in time. The average rate of change of velocity with respect to clock time is the same as the acceleration **
......!!!!!!!!...................................
NOTES ------->
.......................................................!!!!!!!!...................................
20:30:30 If you know average acceleration and time interval what can you find?
......!!!!!!!!...................................
RESPONSE --> I can find the change in velocity when I know the average acceleration and the time interval by multiply the accelaration by the time. confidence assessment: 3
.................................................
......!!!!!!!!...................................
20:33:19 06-25-2008 20:33:19 ** Accel = change in vel / change in clock time, so if you know accel and time interval (i.e., change in clock time) you can find change in vel = accel * change in clock time. In this case you don't know anything about how fast the object is traveling. You can only find the change in its velocity. COMMON ERROR (and response): Average acceleration is the average velocity divided by the time (for the change in the average velocity)so you would be able to find the average velocity by multiplying the average acceleration by the change in time. INSTRUCTOR RESPONSE: Acceleration is rate of change of velocity--change in velocity divided by the change in clock time. It is not average velocity / change in clock time. COUNTEREXAMPLE TO COMMON ERROR: Moving at a constant 60 mph for 3 hours, there is no change in velocity so acceleration = rate of change of velocity is zero. However average velocity / change in clock time = 60 mph / (3 hr) = 20 mile / hr^2, which is not zero. This shows that acceleration is not ave vel / change in clock time. COMMON ERROR and response: You can find displacement INSTRUCTOR RESPONSE: From average velocity and time interval you can find displacement. However from average acceleration and time interval you can find only change in velocity. Acceleration is the rate at which velocity changes so average acceleration is change in velocity/change in clock time. From this it follows that change in velocity = acceleration*change in clock time. **
......!!!!!!!!...................................
NOTES ------->
.......................................................!!!!!!!!...................................
20:33:52 Can you find velocity from average acceleration and time interval?
......!!!!!!!!...................................
RESPONSE --> No we cannot find velocity only average velocity. To find velocity more info is needed. confidence assessment: 3
.................................................
......!!!!!!!!...................................
20:35:22 06-25-2008 20:35:22 ** Ave accel = change in vel / change in clock time. If acceleration is constant, then this relationship becomes acceleration = change in velocity/change in clock time. Change in clock time is the time interval, so if we know time interval and acceleration we can find change in velocity = acceleration * change in clock time = acceleration * change in clock time. We cannot find velocity, only change in velocity. We would need additional information (e.g., initial velocity, average velocity or final velocity) to find an actual velocity. For example if we know that the velocity of a car is changing at 2 (mi/hr) / sec then we know that in 5 seconds the speed will change by 2 (mi/hr)/s * 5 s = 10 mi/hr. But we don't know how fast the car is going in the first place, so we have no information about its actual velocity. If this car had originally been going 20 mi/hr, it would have ended up at 30 miles/hr. If it had originally been going 70 mi/hr, it would have ended up at 80 miles/hr. Similarly if an object is accelerating at 30 m/s^2 (i.e., 30 (m/s) / s) for eight seconds, its velocity will change by 30 meters/second^2 * 8 seconds = 240 m/s. Again we don't know what the actual velocity will be because we don't know what velocity the object was originally moving. ANOTHER SOLUTION: The answer is 'No'. You can divide `ds (change in position) by `dt (change in clock time) to get vAve = `ds / `dt. Or you can divide `dv (change in vel) by `dt to get aAve. So from aAve and `dt you can get `dv, the change in v. But you can't get v itself. EXAMPLE: You can find the change in a quantity from a rate and a time interval, but you can't find the actual value of the quantity. For example, accelerating for 2 sec at 3 mph / sec, your velocity changes by 6 mph, but that's all you know. You don't know how fast you were going in the first place. Could be from 5 mph to 12 mph, or 200 mph to 206 mph (hopefully not down the Interstate). COMMON ERROR: Yes. Final velocity is average velocity multiplied by 2. INSTRUCTOR RESPONSE: We aren't given ave velocity and time interval, we're give ave accel and time interval, so this answer is not valid. Note also that final velocity is average velocity multiplied by 2 ONLY when init vel is zero. Be sure you always state it this way. ANOTHER EXAMPLE: You can't find velocity from ave accel and time interval--you can only find change in velocity from this information. For example a velocity change of 10 mph would result from ave accel 2 m/s^2 for 2 seconds; this change could be between 10 and 20 mph or between 180 and 190 mph, and if all we know is ave accel and time interval we couldn't tell the difference. ONE MORE RESPONSE: You can find the change in velocity. The actual velocity cannot be found from ave accel and time interval. For example you would get the same result for acceleration if a car went from 10 mph to 20 mph in 5 sec as you would if it went from 200 mph to 220 mph in 10 sec. **
......!!!!!!!!...................................
NOTES ------->
.......................................................!!!!!!!!...................................
20:36:10 Can you find change in velocity from average acceleration and time interval?
......!!!!!!!!...................................
RESPONSE --> yes we can find the change in velocity from the average acceleration mulitplied by the time interval. confidence assessment: 3
.................................................
......!!!!!!!!...................................
20:37:15 06-25-2008 20:37:15 **Good student response: Yes, the answer that I provided previously is wrong, I didn't consider the 'change in velocity' I only considered the velocity as being the same as the change in velocity and that was not correct. Change in velocity is average accel * `dt. CALCULUS-RELATED ANSWER WITH INSTRUCTOR NOTE(relevant mostly to University Physics students) Yes, you take the integral with respect to time INSTRUCTOR NOTE: That's essentially what you're doing if you multiply average acceleration by time interval. In calculus terms the reason you can't get actual velocity from acceleration information alone is that when you integrate acceleration you get an arbitrary integration constant. You don't have any information in those questions to evaluate c. **
......!!!!!!!!...................................
NOTES ------->
.......................................................!!!!!!!!...................................
20:37:44 Can you find average velocity from average acceleration and time interval?
......!!!!!!!!...................................
RESPONSE --> Yes by multiplying the average acceleration by the average time interval confidence assessment: 3
.................................................
......!!!!!!!!...................................
20:41:26 06-25-2008 20:41:26 ** CORRECT STATEMENT BUT NOT AN ANSWER TO THIS QUESTION: The average acceleration would be multiplied by the time interval to find the change in the velocity INSTRUCTOR RESPONSE: Your statement is correct, but as you say you can find change in vel, which is not the same thing as ave vel. You cannot find ave vel. from just accel and time interval. There is for example nothing in accel and time interval that tells you how fast the object was going initially. The same acceleration and time interval could apply as well to an object starting from rest as to an object starting at 100 m/s; the average velocity would not be the same in both cases. So accel and time interval cannot determine average velocity. CALCULUS-RELATED ERRONEOUS ANSWER AND INSTRUCTOR CLAIRIFICATION(relevant mostly to University Physics students: Yes, you take the integral and the limits of integration at the time intervals CLARIFICATION BY INSTRUCTOR: A definite integral of acceleration with respect to t gives you only the change in v, not v itself. You need an initial condition to evaluate the integration constant in the indefinite integral. To find the average velocity you would have to integrate velocity (definite integral over the time interval) and divide by the time interval. **
......!!!!!!!!...................................
NOTES ------->
.......................................................!!!!!!!!...................................
20:45:22 You can find only change in velocity from average acceleration and time interval. To find actual velocity you have to know at what velocity you started. Why can't you find average velocity from acceleration and time interval?
......!!!!!!!!...................................
RESPONSE --> I can't find average velocity from acceleration and time interval because average velocity is a rate not a change. Whereas aveerage acceleration is a rate also and requires a change divide by a change, not a rate divided by a change. confidence assessment: 3
.................................................
......!!!!!!!!...................................
20:45:35 06-25-2008 20:45:35 ** Average velocity is change in position/change in clock time. Average velocity has no direct relationship with acceleration. CALCULUS-RELATED ANSWER you dont know the inital velocity or the final velocity INSTRUCTOR COMMENT: . . . i.e., you can't evaluate the integration constant. **
......!!!!!!!!...................................
NOTES ------->
.......................................................!!!!!!!!...................................
20:54:53 General College Physics only: Problem #10 Summarize your solution Problem 1.10 (approx. uncertainty in area of circle given radius 2.8 * 10^4 cm).
......!!!!!!!!...................................
RESPONSE --> 2.5 * 10^9 cm^2 area = pi * r^2 area = 3.14 * (28000^2) area = 246176000 simplified area = 2.5 * 10 ^ 9 cm ^2 confidence assessment: 3
.................................................
......!!!!!!!!...................................
20:56:08 06-25-2008 20:56:08 ** Radius 2.8 * 10^4 cm means that the radius is between 2.75 * 10^4 cm and 2.85 * 10^4 cm. This means that the area is between pi * (2.75 * 10^4 cm)^2 = 2.376 * 10^9 cm^2 and pi * (2.85 * 10^4 cm)^2 = 2.552 * 10^9 cm^2. The difference is .176 * 10^9 cm^2 = 1.76 * 10^8 cm^2, which is the uncertainty in the area. Note that the .1 * 10^4 cm uncertainty in radius is about 4% of the radius, which the .176 * 10^9 cm uncertainty in area is about 8% of the area. This is because the area is proportional to the squared radius. A small percent uncertainty in the radius gives very nearly double the percent uncertainty in the squared radius. **
......!!!!!!!!...................................
NOTES ------->
.......................................................!!!!!!!!...................................
20:57:10 Query Add comments on any surprises or insights you experienced as a result of this assignment.
......!!!!!!!!...................................
RESPONSE --> I did not read into the last question properly in that the uncertainty was to be defined an not the listed solution. confidence assessment: 3
.................................................
......!!!!!!!!...................................
20:57:50 06-25-2008 20:57:50 ** COMMON STUDENT COMMENT AND INSTRUCTOR RESPONSE: I am really confused about velocity, acceleration changes in velocity and acceleration, etc. I guess I am the type that works with a formula and plugs in a number. I have went back to the class notes and the problem sets to summarize formulas. Any suggestions? RESPONSE: I note that you are expressing most of your answers in the form of formulas. Ability to use formulas and plug in numbers is useful, but it doesn't involve understanding the concepts, and without an understanding of the concepts we tend to plug our numbers into equations that don't apply. So we deal first with concepts. However formulas do come along fairly soon. The concepts of velocity, acceleration, etc. are very fundamental, but they are tricky and they take awhile to master. You are doing OK at this point. You'll see plenty more over the next few assignments. If you look at the Linked Outline (on the main Physics 1 page--the one where you click on the Assts button--click on the Overviews button, then on the Linked Outline. You will see a table with a bunch of formulas and links to explanations. You might find this page very useful. Also the Introductory Problem Sets give you formulas in the Generalized Solutions. **
......!!!!!!!!...................................
NOTES ------->
.......................................................!!!!!!!!...................................
21:00:39 Principles of Physics Students and General College Physics Students: Problem 14. What is your own height in meters and what is your own mass in kg, and how did you determine these?
......!!!!!!!!...................................
RESPONSE --> 1.85 m 97.5 kg This is determined by converting feet to meters and pounds to kilograms. Using their appropriate formulas. confidence assessment: 3
.................................................
......!!!!!!!!...................................
21:00:54 06-25-2008 21:00:54 Presumably you know your height in feet and inches, and your weight in pounds. Presumably also, you can convert your height in feet and inches to inches. To get your height in meters, you would first convert your height in inches to cm, using the fact that 1 inch = 2.54 cm. Dividing both sides of 1 in = 2.54 cm by either 1 in or 2.54 cm tells us that 1 = 1 in / 2.54 cm or that 1 = 2.54 cm / 1 in, so any quantity can be multiplied by 1 in / (2.54 cm) or by 2.54 cm / (1 in) without changing its value. Thus if you multiply your height in inches by 2.54 cm / (1 in), you will get your height in cm. For example if your height is 69 in, your height in cm will be 69 in * 2.54 cm / (1 in) = 175 in * cm / in. in * cm / in = (in / in) * cm = 1 * cm = cm, so our calculation comes out 175 cm.
......!!!!!!!!...................................
NOTES ------->
................................................."