course PHY 201 June 14 around 11:30 am ph1 query 1*********************************************
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Given Solution: A rate is a change in something divided by a change in something else. This question concerns velocity, which is the rate of change of position: change in position divided by change in clock time. ** NOTE ON NOTATION Students often quote a formula like v = d / t. It's best to avoid this formula completely. The average velocity on an interval is defined as change in position / change in clock time. • The symbol d doesn't look like a change in anything, nor does the symbol t. • And the symbol v doesn't distinguish between initial velocity, final velocity, average velocity, change in velocity and instantaneous velocity, all of which are important concepts that need to be associated with distinct symbols. In this course we use `d to stand for the capital Greek symbol Delta, which universally indicates the change in a quantity. If we use d for distance, then the 'change in distance' would be denoted `dd. Very confusing to have two different d's, with two different meanings, in the same expression. We generally use s or x to stand for position, so `ds or `dx would stand for change in position. Change in clock time would be `dt. Thus v_Ave = `ds / `dt (or alternatively, if we use x for position, v_Ave = `dx / `dt). With this notation we can tell that we are dividing change in position by change in clock time. For University Physics students (calculus-based note): If x is the position then velocity is dx/dt, the derivative of position with respect to clock time. This is the limiting value of the rate of change of position with respect to clock time. You need to think in these terms. v stands for instantaneous velocity. v_Ave stands for the average velocity on an interval. If you used d for position then you would have the formula v = dd / dt. The dd in the numerator doesn't make a lot of sense; one d indicates the infinitesimal change in the other d. &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): OK ********************************************* Question: Given average speed and time interval how do you find distance moved? YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: When given the average speed and time required, you can switch the equation around and solve for the changed distance. Ave speed = change in distance / change in time. So, you switch it around to solve for change in distance: change in distance = ave. speed * change in time. confidence #$&* 3 ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: ** You multiply average speed * time interval to find distance moved. For example, 50 miles / hour * 3 hours = 150 miles. ** &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): 3 ********************************************* Question: Given average speed and distance moved how do you find the corresponding time interval? YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: If you are given average speed and distance moved you can find the clock time it took: average speed = distance moved / time interval. So, you switch the equation around to solve for time interval: time interval = distance moved / average speed. confidence #$&* 3 ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: ** time interval = distance / average speed. For example if we travel 100 miles at 50 mph it takes 2 hours--we divide the distance by the speed. In symbols, if `ds = vAve * `dt then `dt = `ds/vAve. Also note that (cm/s ) / s = cm/s^2, not sec, whereas cm / (cm/s) = cm * s / cm = s, as appropriate in a calculation of `dt. ** &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): OK ********************************************* Question: Given time interval and distance moved how do you get average speed? YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: If you are given the time and distance you can easily find the average speed: average speed = distance moved / time interval. confidence #$&* 3 ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: ** Average speed = distance / change in clock time. This is the definition of average speed. For example if we travel 300 miles in 5 hours we have been traveling at an average speed of 300 miles / 5 hours = 60 miles / hour. ** &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): OK ********************************************* Question: A ball rolls from rest down a book, off that book and onto another book, where it picks up speed before rolling off the end of that book. Consider the interval that begins when the ball first encounters the second book, and ends when it rolls of the end of the book. For this interval, place in order the quantities initial velocity (which we denote v_0), and final velocity (which we denote v_f), average velocity (which we denote v_Ave). During this interval, the ball's velocity changes. It is possible for the change in its velocity to exceed the three quantities you just listed? Is it possible for all three of these quantities to exceed the change in the ball's velocity? Explain. Note that the change in the ball's velocity is denoted `dv. YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: For this situation I am going to use an example and then explain the experiment: For the interval let’s say the initial velocity is v_5, the final velocity v_2, and the average velocity 3.5. The change in its velocity, ‘dv, cannot exceed the three quantities. It will exceed the final velocity, because at the end the ball is traveling fast and yet covering more area. No because the ending point has to be smaller than the ball’s change in velocity, because of the speed the ball is traveling at. confidence #$&* 2 ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^ ********************************************* Question: If the position of an object change be 5.2 meters, with an uncertainty of +-4%, during a time interval of 1.3 seconds, with an uncertainty of +-2%, then What is the uncertainty in the change in position in meters> What is the uncertainty in the time interval in seconds? What is the average velocity of the object, and what do you think is the uncertainty in the average velocity? (this last question is required of University Physics students only, but other are welcome to answer): What is the percent uncertainty in the average velocity of the object, and what is the uncertainty as given in units of velocity? YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: To the uncertainty in the change in position you multiply the meters by the percent: 5.2 m * .04 = .208 is the level of uncertainty. The uncertainty in the time interval of seconds is: 1.3 sec *.02 = .026 is the level of uncertainty. The average velocity is the distance change divided by the time change: average velocity = 5.2 m / 1.3 sec = 4 m per sec. I think the uncertainty of the average velocity is the two levels of uncertainty subtracted together: .208 - .026 = .182 is the level of uncertainty for the average velocity. The percent uncertainty for the average velocity is: .182 / 4 m per sec = .0455 * 100 = 4.55% uncertainty. confidence #$&* 3 ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^