#$&* course phy 231 June 27 8:38 pm 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: To find the distance an object has moved given average speed and a time internal, take average speed and multiply it by the time interval to cover that distance. distance = (ave speed) * (time interval) confidence rating #$&*:: 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):OK ********************************************* Question: Given average speed and distance moved how do you find the corresponding time interval? YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: To find the time interval take the distance moved and divide it by the average speed. confidence rating #$&*:: 3 ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: ** time interval = distance / average speed. For example if we travel 100 miles at 50 mph it takes 2 hours--we divide the distance by the speed. In symbols, if `ds = vAve * `dt then `dt = `ds/vAve. Also note that (cm/s ) / s = cm/s^2, not sec, whereas cm / (cm/s) = cm * s / cm = s, as appropriate in a calculation of `dt. ** &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): In a previous problem on another assignment I incorrectly put the units of change in velocity/time in cm instead of cm/s^2, I understand my mistake as well that cm/(cm/s) = seconds. Question: Given time interval and distance moved how do you get average speed? Your solution: To determine the average speed, take the distance moved and divide it by the time interval. confidence rating #$&*:: 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: In order, v_0, v_Ave, v_f The final velocity should be the greatest. The final velocity of a ball rolling down an incline would always be the greatest value in comparison to the average velocity. The average velocity is going to include in the average, the extreme values to find the “average.” This average will always be less than the greatest extreme, which in the experiment with a ball rolling down an incline; the greatest extreme would be the final velocity or v_f. The change in the ball’s velocity, `dv, would also never exceed the v_f. confidence rating #$&*: 3 ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^ ********************************************* Question: If the velocity at the beginning of an interval is 4 m/s and at the end of the interval it is 10 m/s, then what is the average of these velocities, and what is the change in velocity? List the four quantities initial velocity, final velocity, average of initial and final velocities, and change in velocity, in order from least to greatest. Give an example of positive initial and final velocities for which the order of the four quantities would be different. For positive initial and final velocities, is it possible for the change in velocity to exceed the other three quanities? YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: The average velocity given 4 m/s and 10 m/s would be (4 m/s + 10 m/s)/2 = 7 m/s. The change in velocity is 10 m/s - 4 m/s = 6 m/s. In order from least to greatest: Initial v_0, Change `dv, Average v_Ave, Final v_f If the velocity at the beginning of an interval is 10 m/s and at the end 4 m/s, the order from least to greatest of initial, average, change and final velocity would be: Final 4 m/s Change 6 m/s Average 7 m/s Initial 10 m/s confidence rating #$&*:3 ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^ ********************************************* Question: If the position of an object changes by 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 ia 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: The uncertainty of the change in position is .04 of 5.2 = +-0.216, so the position could be between 5.416 meters and 5.184 meters. The uncertainty of the time interval is .02 of 1.3 = +-0.026, so the time could be between 1.274 sec and 1.326 sec. The average velocity of the object is 5.2 m/1.3 sec = 4 m/sec The uncertainty in the average velocity is calculated by finding the average velocity at 5.416 m and 1.274 sec, the largest distance and the quickest time. v ave = 5.416 m/1.274 sec = 4.25 m/sec. Calculate the average velocity again, this time using 5.184 m/1.326 sec = 3.91 m/sec. These two values give you the extreme limits of average velocity due to the uncertainty of distance and time. 3.91/4 = .9775 or 4.25/4 = 1.06 therefore the resulting average velocities differ between about 2% and 6% from 4 m/sec. So the uncertainty for the average velocity is 8% every m/sec. confidence rating #$&*: ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^ I felt very confident about my answer until the last question. Not sure if I calculated correctly the uncertainty for the average velocity. " Self-critique (if necessary): ------------------------------------------------ Self-critique rating: