#$&* course Phy 201 2/2/13 at 9:21 PM ph1 query 1 x
.............................................
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 the average rate of change of position with respect to clock time. By the definition of average rate, then, the average velocity on the interval is v_ave = (change in position / change in clock time). One reason we might not want to use v = d / t: The symbol d doesn’t look like a change in anything, nor does the symbol t. Also it’s very to read ‘d’ and ‘distance’ rather than ‘displacement’. Another reason: 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. It’s potentially 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): I said that quantity A would be the change in distance and the answer said that it was the change in position. ???? Does these two statements mean the same thing???? ------------------------------------------------ Self-critique rating:
.............................................
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 ------------------------------------------------ Self-critique rating: OK #$&* ********************************************* Question: Given average speed and distance moved how do you find the corresponding time interval? YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: If you are given the average speed and distance moved, then you can find the corresponding time interval by dividing the distance moved by the average speed. confidence rating #$&*: ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^ 3
.............................................
Given Solution: ** time interval = distance / average speed. For example if we travel 100 miles at 50 mph it takes 2 hours--we divide the distance by the speed. In symbols, if `ds = vAve * `dt then `dt = `ds/vAve. Also note that (cm/s ) / s = cm/s^2, not sec, whereas cm / (cm/s) = cm * s / cm = s, as appropriate in a calculation of `dt. ** &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): OK ------------------------------------------------ Self-critique rating: OK #$&* ********************************************* Question: Given time interval and distance moved how do you get average speed? YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: If you are given the time interval and distance moved, then you can find the average speed by dividing the distance moved by the time interval. confidence rating #$&*: ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^ 3
.............................................
Given Solution: ** Average speed = distance / change in clock time. This is the definition of average speed. For example if we travel 300 miles in 5 hours we have been traveling at an average speed of 300 miles / 5 hours = 60 miles / hour. ** &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): OK ------------------------------------------------ Self-critique rating: OK #$&* ********************************************* 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 interval the quantities in order would be the initial velocity, average velocity, and then the final velocity. No, because the change in velocity can not exceed the final velocity in this interval. It could be possible for the change in velocity to exceed the initial velocity and average velocity if the initial velocity was very small and the final velocity was very big, but it would not exceed the final velocity. An example could be: Initial velocity = 1 m/s, Average velocity = 10.5 m/s, Final velocity = 20 m/s so from this the change in velocity would be 20 - 1 =19 m/s so the change in velocity would be bigger than the initial velocity and the average velocity, but not the final velocity. Yes, it is possible for all three of these quantities to exceed the change in the ball’s velocity. One way this could happen is if the initial velocity was pretty big to start with and then the final velocity wasn’t much higher than that then all three of those quantities could exceed the change in the ball’s velocity. An example could be: Initial velocity = 30 m/s, Average velocity = 35 m/s, Final velocity = 40 m/s so from this the Change in velocity would be 40 - 30 = 10 m/s which is smaller than all three of those quantities. 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: Average velocity = 4 m/s + 10 m/s = 14 m/s Average velocity = 14 m/s / 2 = 7 m/s Average velocity = 7 m/s The change in velocity is 10 m/s - 4 m/s = 6 m/s The four quantities in order from least to greatest is: Initial velocity 4 m/s, Change in velocity 6 m/s, Average velocity 7 m/s, Final velocity 10 m/s An example of positive initial and final velocities for which the order of the four quantities would be different could be Initial velocity = 30 m/s, Final velocity = 40 m/s so from this the Change in velocity would be 40 - 30 = 10 m/s, and the Average velocity would be (30 + 40) / 2 = 35 m/s Given this example, the order from least to greatest would be: Change in velocity 10 m/s, Initial velocity 30 m/s, Average velocity 35 m/s, Final velocity 40 m/s No, because the change in velocity can not exceed the final velocity if the initial and final velocities are positive. It could be possible for the change in velocity to exceed the initial velocity and average velocity if the initial velocity was very small and the final velocity was very big, but it would not exceed the final velocity. An example could be: Initial velocity = 1 m/s, Average velocity = 10.5 m/s, Final velocity = 20 m/s so from this the change in velocity would be 20 - 1 =19 m/s so the change in velocity would be bigger than the initial velocity and the average velocity, but not the final velocity. confidence rating #$&*: ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^ 3 #$&* ********************************************* Question: If the position of an object changes by 5.2 meters, with an uncertainty of +-4%, during a time interval of 1.3 seconds, with an uncertainty of +-2%, then What is the uncertainty in the change in position in meters> What is the uncertainty in the time interval in seconds? What is the average velocity of the object, and what do you think ia the uncertainty in the average velocity? (this last question is required of University Physics students only, but other are welcome to answer): What is the percent uncertainty in the average velocity of the object, and what is the uncertainty as given in units of velocity? YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: The uncertainty in the change in position in meters is = + or - .04 (or 4%). So the change in the position is probably between 5.16 to 5.24. The percent uncertainty = (.04 / 5.2) * 100 = .77% The uncertainty in the time interval in seconds is = + or - .02 (or 2%). So the time interval is probably between 1.28 to 1.32. The percent uncertainty = (.02/1.3) * 100 = 1.5% The average velocity of the object is 5.2 meters / 1.3 seconds = 4 meters/second I think the uncertainty in the average velocity would be (+ or - .04) + (+ or - .02) = + or - .06 (or 6%). The percent uncertainty would be = (.06 / 4) * 100 = 1.5% confidence rating #$&*: ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^ 2 #$&* " Self-critique (if necessary): ------------------------------------------------ Self-critique rating: OK #$&* ********************************************* 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 interval the quantities in order would be the initial velocity, average velocity, and then the final velocity. No, because the change in velocity can not exceed the final velocity in this interval. It could be possible for the change in velocity to exceed the initial velocity and average velocity if the initial velocity was very small and the final velocity was very big, but it would not exceed the final velocity. An example could be: Initial velocity = 1 m/s, Average velocity = 10.5 m/s, Final velocity = 20 m/s so from this the change in velocity would be 20 - 1 =19 m/s so the change in velocity would be bigger than the initial velocity and the average velocity, but not the final velocity. Yes, it is possible for all three of these quantities to exceed the change in the ball’s velocity. One way this could happen is if the initial velocity was pretty big to start with and then the final velocity wasn’t much higher than that then all three of those quantities could exceed the change in the ball’s velocity. An example could be: Initial velocity = 30 m/s, Average velocity = 35 m/s, Final velocity = 40 m/s so from this the Change in velocity would be 40 - 30 = 10 m/s which is smaller than all three of those quantities. 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: Average velocity = 4 m/s + 10 m/s = 14 m/s Average velocity = 14 m/s / 2 = 7 m/s Average velocity = 7 m/s The change in velocity is 10 m/s - 4 m/s = 6 m/s The four quantities in order from least to greatest is: Initial velocity 4 m/s, Change in velocity 6 m/s, Average velocity 7 m/s, Final velocity 10 m/s An example of positive initial and final velocities for which the order of the four quantities would be different could be Initial velocity = 30 m/s, Final velocity = 40 m/s so from this the Change in velocity would be 40 - 30 = 10 m/s, and the Average velocity would be (30 + 40) / 2 = 35 m/s Given this example, the order from least to greatest would be: Change in velocity 10 m/s, Initial velocity 30 m/s, Average velocity 35 m/s, Final velocity 40 m/s No, because the change in velocity can not exceed the final velocity if the initial and final velocities are positive. It could be possible for the change in velocity to exceed the initial velocity and average velocity if the initial velocity was very small and the final velocity was very big, but it would not exceed the final velocity. An example could be: Initial velocity = 1 m/s, Average velocity = 10.5 m/s, Final velocity = 20 m/s so from this the change in velocity would be 20 - 1 =19 m/s so the change in velocity would be bigger than the initial velocity and the average velocity, but not the final velocity. confidence rating #$&*: ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^ 3 #$&* ********************************************* Question: If the position of an object changes by 5.2 meters, with an uncertainty of +-4%, during a time interval of 1.3 seconds, with an uncertainty of +-2%, then What is the uncertainty in the change in position in meters> What is the uncertainty in the time interval in seconds? What is the average velocity of the object, and what do you think ia the uncertainty in the average velocity? (this last question is required of University Physics students only, but other are welcome to answer): What is the percent uncertainty in the average velocity of the object, and what is the uncertainty as given in units of velocity? YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: The uncertainty in the change in position in meters is = + or - .04 (or 4%). So the change in the position is probably between 5.16 to 5.24. The percent uncertainty = (.04 / 5.2) * 100 = .77% The uncertainty in the time interval in seconds is = + or - .02 (or 2%). So the time interval is probably between 1.28 to 1.32. The percent uncertainty = (.02/1.3) * 100 = 1.5% The average velocity of the object is 5.2 meters / 1.3 seconds = 4 meters/second I think the uncertainty in the average velocity would be (+ or - .04) + (+ or - .02) = + or - .06 (or 6%). The percent uncertainty would be = (.06 / 4) * 100 = 1.5% confidence rating #$&*: ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^ 2 #$&* " Self-critique (if necessary): ------------------------------------------------ Self-critique rating: #*&!