course Phy231 6/10 3:45 004. `Query 4NOTE PRELIMINARY TO QUERY:
.............................................
Given Solution: ** Velocity is the rate of change of position with respect to clock time. Acceleration is rate of change of velocity with respect to clock time. 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 clock time. note that the term 'average rate of change of velocity with respect to clock time' means the same thing as 'acceleration' ** Self-critique (if necessary): OK Self-critique rating #$&* OK Question: If you know average acceleration and time interval what can you find? Your solution: You can find the change in velocity by multiplying the average acceleration by the time interval. confidence rating #$&*: 3 ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
.............................................
Given Solution: ** 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. Self-critique (if necessary): OK Self-critique rating #$&* OK Question: Can you find velocity from average acceleration and time interval? Your solution: No, you can only find the change in velocity. The equation is aAve = dv/dt. You can only solve for dv, but you cannot get just v. confidence rating #$&*: 2 ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
.............................................
Given Solution: ** 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. 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. &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): OK Self-critique rating #$&* OK Question: `qCan you find change in velocity from average acceleration and time interval? Your solution: Yes, you multiply the average acceleration and the time interval to get the change in velocity. confidence rating #$&*: 3 ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
.............................................
Given Solution: Average acceleration is ave rate of change of velocity with respect to clock time, which is `dv / `dt. Given average acceleration and time interval you therefore know aAve = `dv / `dt, and you know `dt. The obvious use of these quantities is to multiply them: aAve * `dt = `dv / `dt * `dt = `dv So with the given information aAve and `dt, we can find `dv, which is the change in velocity. From this information we can find nothing at all about the average velocity vAve, which is a quantity which is completely unrelated to `dv . 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 v. IMPORTANT INSTRUCTOR NOTE: Always modify the term 'velocity' or the symbol 'v'. I do not use v or the unmodified term 'velocity' for anything at this stage of the course, and despite the fact that your textbook does, you should at this stage consider avoiding it as well. At this point in the course the word 'velocity' should always be modified by an adjective. Motion on any interval involves the following quantities, among others: initial velocity v0, the velocity at the beginning of the interval final velocity vf, the velocity at the end of the interval average velocity vAve, defined as average rate of change of position with respect to clock time, `ds / `dt change in velocity `dv, which is the difference between initial and final velocities (midpoint velocity vMid), which is the same as vAve provided the v vs. t graph is linear (i.e., provided acceleration is constant); since most motion problems will involve uniform acceleration this quantity will be seen than the others If you aren't specific about which velocity you mean, you will tend to confuse one or more of these quantities. The symbol v, and the unmodified term 'velocity', have more complex and ambiguous meanings than the specific terms outlined above: The symbol v stands for 'instantaneous velocity', a concept that is challenging to understand well without a calculus background (which isn't expected or required for your the General College Physics or Principles of Physics courses). Your text (along with most others) uses v to stand for the instantaneous velocity at clock time t, but sometimes it uses the symbol v for the average velocity. To denote an instantanous velocity I consider it more appropriate to use the functional notation v(t)., which clearly denotes the velocity at a specific instant. The ambiguous use of the word 'velocity' and the symbol 'v' are the source of almost universal confusion among students in non-calculus-based physics courses. (Students in calculus-based courses are expected to have the background to understand these distinctions, though most such students can also profit from the specific terminology outlined here. Self-critique (if necessary): OK Self-critique rating #$&* OK Question: `qCan you find average velocity from average acceleration and time interval? Your solution: No, you cannot get the average velocity from the aAve and the time interval. You do not have the initial or final velocities to determine what the average was. confidence rating #$&*: 2 ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
.............................................
Given Solution: 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. 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. ** Self-critique (if necessary): You only would need the initial velocity to determine what the average was; the final velocity is not required. Self-critique rating #$&* 3 Question: `qYou can find only change in velocity from average acceleration and time interval. To find actual velocity you have to know at what velocity you started. Why can't you find average velocity from acceleration and time interval? Your solution: You do not have the initial velocity or the final velocity to integrate. Also average acceleration does not have anything to do with average velocity. confidence rating #$&*: 2 ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
.............................................
Given Solution: ** 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 . . . i.e., you can't evaluate the integration constant. ** Self-critique (if necessary): OK Self-critique rating #$&* OK Question: Give at least three possible units for velocity, and at least three possible units for clock time. Give at least three possible units for the slope between two points of a graph of velocity vs. clock time. Explain how you reasoned out the answer to this question. Explain the meaning of the slope of this graph. YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: Velocity: m/s, cm/s, mph Clock time: seconds, hours, minutes Slope: m/s^2, cm/s^2, km/hr^2 The slope of the velocity graph is going to be the velocity units divided by the clock time units, so you get m/s / s = m/s*s = m/s^2. Also, the slope of the graph is the same thing as the average acceleration, so the units will be the acceleration units. confidence rating #$&*: 3 ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
.............................................
Given Solution: Possible units for velocity might include millimeters / hour, kilometers / second, or meters / minute. The standard unit is meters / second. Possible units for clock time might include microseconds, minutes, years. The standard unit is the second. The slope between two points of a graph is the rise of the graph divided by its run. The rise between two points of a graph of velocity vs. clock time represents the change in the velocity between these points. So the rise might have units of, say, millimeters / hour or meters / minute. The standard unit would be meters / second. The run between two points of a graph of velocity vs. clock time represents the change in the clock time between these points. So the run might have units of, say, microseconds or years. The standard unit is the second. The units of slope are units of rise divided by units of run. So the units of the slope might be any of the following: (millimeters / hour) / microsecond, which by the rules for multiplying and dividing fractions would simplify to (millimeters / hour) * (1 / microsecond) = millimeters / (hour * microsecond). (meters / minute) / year, which by the rules for multiplying and dividing fractions would simplify to (meters / minute) * (1 / year) = meters / (minute * year) or the standard unit, (meters / second) / second, which by the rules for multiplying and dividing fractions would simplify to to (meters / second) * (1 / second) = meters / (second * second) = meters / second^2. Note that a unit like millimeters / (hour * microsecond) could be converted to standard units. Since 1000 millimeters = 1 meter we can use conversion factor (1000 millimeters) / (1 meter) or (1 meter) / (1000 millimeters) Since 1 hour = 3600 seconds we can use conversion factor (1 hour) / (3600 seconds), or (3600 seconds) / (1 hour) Since 10^6 microseconds = 1 second we have conversion factors (10^6 microseconds) / (1 second) and (1 second) / (10^6 microseconds). If we understand the rules for fractions, we can easily apply these conversion factors to get the following: millimeters / (hour * microsecond) = mm / (hr microsec) * 1 m / (1000 mm) * 1 hr / (3600 sec) * 10^6 microsec / (1 sec) = mm * hr * 10^6 microsec m/ (hr microsec mm * 3600 sec * sec) = (10^6 / 3600) * (mm hr microsec m) / (hr microsec mm sec sec) = (10^6 / 3600) * m / sec^2 &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): You should convert the units into standard units to make it easier ti understand/visualize. Self-critique rating #$&* 3 ********************************************* Question: If the velocity of an object changes at a uniform rate from 5 m/s to 13 m/s between clock times t = 7 s and t = 11 s then on this interval what is its average velocity, and what is the average rate at which its velocity changes with respect to clock time? YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: aAve = 13m/s-5m/s / 11s-5s = 8m/s / 6s = 1.33m/s^2 confidence rating #$&*: 3 ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^ ********************************************* Question: Explain how to solve the relationship aAve = `dv / `dt for `dt. YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: You cross multiply the proportion and get aAve*dt = dv. Then you divide both sides by aAve to isolate the dt. Therefore, dt = dv/aAve. confidence rating #$&*: 3 ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
.............................................
Given Solution: &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): Im fairly certain I got that right Self-critique rating #$&* OK Question for University Physics Students: What is the instantaneous rate of change of v with respect to t at t = 2, given that v(t) = 2 t^2 - t + 3? Explain how you obtained this result. What is the expression for the instantaneous rate of change of v with respect to t at general clock time t, given the same velocity function? Explain how you determined this. Your solution: V(t) = 4t-1 at t=2 v(2) = 4(2)-1 = 7 distance/time/time You take the derivative of the velocity function, which gives you the acceleration function. Then you substitute in the clock time and simplify, which gives you the instantaneous acceleration at that point. The general expression for the instantaneous rate of change of velocity is v(t)=4t-1. You take the derivative, but do not substitute anything in. confidence rating #$&*: 3 ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
.............................................
Given Solution: v(2) = 2 * 2^2 - 2 + 3 = 9. v(2.1) = 2 * (2.1)^2 - 2.1 + 3 = 9.72 So the average rate of change of v with respect to t for the interval from t = 2 to t = 2.1 is ave rate = change in v / change in t = (9.72 - 9) / (2.1 - 2) = 7.2 v(2.01) = 2.0702, and v(2.001) = 2.007002. Using these values along with v(2) = 9 we find that on interval t = 2 to 2 = 2.01 the average rate is 7.02 on interval t = 2 to 2 = 2.001 the average rate is 7.002 It is therefore reasonable to conjecture that the instantaneous rate at t = 2 is exactly 7. In fact the instantaneous rate of change function is the derivative function v ' (t) = dv / dt = 4 t - 1. This function gives the instantaneous acceleration at clock time t: a(t) = v ' (t) = 4 t - 1. Evaluating this function at t = 2 we obtain a(2) = 4 * 2 - 1 = 8 - 1 = 7, which confirms the conjecture we make based on the series of intervals above. Self-critique (if necessary): I just went ahead and took the derivative for the first part. Is that okay?
.............................................
Given Solution: `a** Student Response: I think I really confused what information stood for what in the Force and Pendulum Experiment. However, I enjoy doing the flow diagrams. They make you think in a different way than you are used to. INSTRUCTOR NOTE: These diagrams are valuable for most people. Not all--it depends on learning style--but most. ** Comments: None "