#$&* course Phy 121 2/15 12:20 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: In order to find distance moved when given average speed and time interval, we would multiply the two. For example, if an object is moving at 50 miles per hour for 2 hours, it has traveled a total distance of 100 miles. This is found by multiplying the avg. speed by the time: 50mph * 2 hours = 100 miles. 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: If we are given the average speed and distance moved, we can find the corresponding time interval by dividing the distance by the average speed. For example, if we had a distance of 200 miles and an average speed of 20 miles per hour we would get a time interval of 10 hours. We could find it the following way: 200 miles / 20 mph = 10 hours. confidence rating #$&*: 3 ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: ** time interval = distance / average speed. For example if we travel 100 miles at 50 mph it takes 2 hours--we divide the distance by the speed. In symbols, if `ds = vAve * `dt then `dt = `ds/vAve. Also note that (cm/s ) / s = cm/s^2, not sec, whereas cm / (cm/s) = cm * s / cm = s, as appropriate in a calculation of `dt. ** &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): OK ********************************************* Question: Given time interval and distance moved how do you get average speed? YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: If we have a given time interval and distance moved, we divide the distance moved by the time interval to get an average speed. For example if we had a distanced moved of 250 miles and a time interval of 5 hours, we would have an average speed of 50 miles per hour. 250 miles / 5 hours = 50 miles per hour. confidence rating #$&*: 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: As long as the velocity changes in a consistent positive manner, then it would not exceed all of the three quantities listed. It is possible that if the ‘dv is fairly quick, it could exceed the v_0 but not the other two quantities. However, if there was a quick spike in the change of the balls velocity, and then it slowed down, that small amount of ‘dv could be bigger than all of the quantities. It is possible for the v_0, v_f, and v_Ave to all exceed the ‘dv. In our example, the initial velocity is not zero because the ball is already rolling when the initial velocity on the second book starts. Therefore, if the ‘dv was consistent, but slow, then it could be less than the three given quantities. confidence rating #$&*: 2 ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^ ********************************************* Question: If the velocity at the beginning of an interval is 4 m/s and at the end of the interval it is 10 m/s, then what is the average of these velocities, and what is the change in velocity? List the four quantities initial velocity, final velocity, average of initial and final velocities, and change in velocity, in order from least to greatest. Give an example of positive initial and final velocities for which the order of the four quantities would be different. For positive initial and final velocities, is it possible for the change in velocity to exceed the other three quanities? YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: The average of the given velocities (4m/s and 10m/s) is 7 m/s. The change in velocity is 6 m/s. The four quantities in order from least to greatest are as follows: v_0, ‘dv, v_Ave, v_f. An example of positive initial and final velocities in which the four quantities would be different could be if a ball was slowing down from initial velocity to final velocity. This would cause the final velocity to be the least and the initial velocity to be the greatest. If the initial and final velocities are both positive, I don’t believe it would be possible for the change in velocity to exceed all three of the other quantities. Although the change in velocity could definitely exceed the average velocity, the initial and final velocities would be a different story. The change in velocity could exceed either the initial or final velocities, but not both at the same time if they both are positive. confidence rating #$&*: 2 ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^ ********************************************* 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 +- 4% or +- .2 The uncertainty in the time interval in seconds is +- 2% or +- .02 The average velocity is calculated by finding the change in distance divided by the time elapsed. 5.2 meters/1.3 seconds = 4 meters per second. I believe the uncertainty in the velocity is only as accurate as the least accurate part, which was the distance. So the velocity is accurate to an uncertainty of +-.2 or +-5%. confidence rating #$&*: 2 ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^ " Self-critique (if necessary): ------------------------------------------------ Self-critique rating:
#$&* course Phy 121 2/15 12:20 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: In order to find distance moved when given average speed and time interval, we would multiply the two. For example, if an object is moving at 50 miles per hour for 2 hours, it has traveled a total distance of 100 miles. This is found by multiplying the avg. speed by the time: 50mph * 2 hours = 100 miles. confidence rating #$&*: 3 ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: ** You multiply average speed * time interval to find distance moved. For example, 50 miles / hour * 3 hours = 150 miles. ** &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): OK ********************************************* Question: Given average speed and distance moved how do you find the corresponding time interval? YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: If we are given the average speed and distance moved, we can find the corresponding time interval by dividing the distance by the average speed. For example, if we had a distance of 200 miles and an average speed of 20 miles per hour we would get a time interval of 10 hours. We could find it the following way: 200 miles / 20 mph = 10 hours. confidence rating #$&*: 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): OK ********************************************* Question: Given time interval and distance moved how do you get average speed? YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: If we have a given time interval and distance moved, we divide the distance moved by the time interval to get an average speed. For example if we had a distanced moved of 250 miles and a time interval of 5 hours, we would have an average speed of 50 miles per hour. 250 miles / 5 hours = 50 miles per hour. 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: As long as the velocity changes in a consistent positive manner, then it would not exceed all of the three quantities listed. It is possible that if the ‘dv is fairly quick, it could exceed the v_0 but not the other two quantities. However, if there was a quick spike in the change of the balls velocity, and then it slowed down, that small amount of ‘dv could be bigger than all of the quantities. It is possible for the v_0, v_f, and v_Ave to all exceed the ‘dv. In our example, the initial velocity is not zero because the ball is already rolling when the initial velocity on the second book starts. Therefore, if the ‘dv was consistent, but slow, then it could be less than the three given quantities. confidence rating #$&*: 2 ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^ ********************************************* 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 of the given velocities (4m/s and 10m/s) is 7 m/s. The change in velocity is 6 m/s. The four quantities in order from least to greatest are as follows: v_0, ‘dv, v_Ave, v_f. An example of positive initial and final velocities in which the four quantities would be different could be if a ball was slowing down from initial velocity to final velocity. This would cause the final velocity to be the least and the initial velocity to be the greatest. If the initial and final velocities are both positive, I don’t believe it would be possible for the change in velocity to exceed all three of the other quantities. Although the change in velocity could definitely exceed the average velocity, the initial and final velocities would be a different story. The change in velocity could exceed either the initial or final velocities, but not both at the same time if they both are positive. confidence rating #$&*: 2 ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^ ********************************************* Question: If the position of an object changes by 5.2 meters, with an uncertainty of +-4%, during a time interval of 1.3 seconds, with an uncertainty of +-2%, then What is the uncertainty in the change in position in meters> What is the uncertainty in the time interval in seconds? What is the average velocity of the object, and what do you think ia the uncertainty in the average velocity? (this last question is required of University Physics students only, but other are welcome to answer): What is the percent uncertainty in the average velocity of the object, and what is the uncertainty as given in units of velocity? YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: The uncertainty in the change in position in meters is +- 4% or +- .2 The uncertainty in the time interval in seconds is +- 2% or +- .02 The average velocity is calculated by finding the change in distance divided by the time elapsed. 5.2 meters/1.3 seconds = 4 meters per second. I believe the uncertainty in the velocity is only as accurate as the least accurate part, which was the distance. So the velocity is accurate to an uncertainty of +-.2 or +-5%. confidence rating #$&*: 2 ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^ " Self-critique (if necessary): ------------------------------------------------ Self-critique rating: #*&!
#$&* course Phy 121 2/15 12:20 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: In order to find distance moved when given average speed and time interval, we would multiply the two. For example, if an object is moving at 50 miles per hour for 2 hours, it has traveled a total distance of 100 miles. This is found by multiplying the avg. speed by the time: 50mph * 2 hours = 100 miles. 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: If we are given the average speed and distance moved, we can find the corresponding time interval by dividing the distance by the average speed. For example, if we had a distance of 200 miles and an average speed of 20 miles per hour we would get a time interval of 10 hours. We could find it the following way: 200 miles / 20 mph = 10 hours. confidence rating #$&*: 3 ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: ** time interval = distance / average speed. For example if we travel 100 miles at 50 mph it takes 2 hours--we divide the distance by the speed. In symbols, if `ds = vAve * `dt then `dt = `ds/vAve. Also note that (cm/s ) / s = cm/s^2, not sec, whereas cm / (cm/s) = cm * s / cm = s, as appropriate in a calculation of `dt. ** &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): OK ********************************************* Question: Given time interval and distance moved how do you get average speed? YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: If we have a given time interval and distance moved, we divide the distance moved by the time interval to get an average speed. For example if we had a distanced moved of 250 miles and a time interval of 5 hours, we would have an average speed of 50 miles per hour. 250 miles / 5 hours = 50 miles per hour. confidence rating #$&*: 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: As long as the velocity changes in a consistent positive manner, then it would not exceed all of the three quantities listed. It is possible that if the ‘dv is fairly quick, it could exceed the v_0 but not the other two quantities. However, if there was a quick spike in the change of the balls velocity, and then it slowed down, that small amount of ‘dv could be bigger than all of the quantities. It is possible for the v_0, v_f, and v_Ave to all exceed the ‘dv. In our example, the initial velocity is not zero because the ball is already rolling when the initial velocity on the second book starts. Therefore, if the ‘dv was consistent, but slow, then it could be less than the three given quantities. confidence rating #$&*: 2 ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^ ********************************************* Question: If the velocity at the beginning of an interval is 4 m/s and at the end of the interval it is 10 m/s, then what is the average of these velocities, and what is the change in velocity? List the four quantities initial velocity, final velocity, average of initial and final velocities, and change in velocity, in order from least to greatest. Give an example of positive initial and final velocities for which the order of the four quantities would be different. For positive initial and final velocities, is it possible for the change in velocity to exceed the other three quanities? YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: The average of the given velocities (4m/s and 10m/s) is 7 m/s. The change in velocity is 6 m/s. The four quantities in order from least to greatest are as follows: v_0, ‘dv, v_Ave, v_f. An example of positive initial and final velocities in which the four quantities would be different could be if a ball was slowing down from initial velocity to final velocity. This would cause the final velocity to be the least and the initial velocity to be the greatest. If the initial and final velocities are both positive, I don’t believe it would be possible for the change in velocity to exceed all three of the other quantities. Although the change in velocity could definitely exceed the average velocity, the initial and final velocities would be a different story. The change in velocity could exceed either the initial or final velocities, but not both at the same time if they both are positive. confidence rating #$&*: 2 ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^ ********************************************* 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 +- 4% or +- .2 The uncertainty in the time interval in seconds is +- 2% or +- .02 The average velocity is calculated by finding the change in distance divided by the time elapsed. 5.2 meters/1.3 seconds = 4 meters per second. I believe the uncertainty in the velocity is only as accurate as the least accurate part, which was the distance. So the velocity is accurate to an uncertainty of +-.2 or +-5%. confidence rating #$&*: 2 ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^ " Self-critique (if necessary): ------------------------------------------------ Self-critique rating: #*&!#*&!