#$&* course Phy 231 delim #$&**********************************************
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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 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): Is ""clock time"" specified to mean the time you read on the clock at a particular instant, whereas ""time"" could indicate clock time or elapsed time? Is it important to use the term ""clock time"" in this way? ------------------------------------------------ Self-critique rating: 2 #$&* ********************************************* Question: Given average speed and time interval how do you find distance moved? YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: We can rearrange the formula above. Average velocity equals change in position divided by change in clock time, or: v_Ave=(`ds)/(`dt) We are looking for change in position, so we rearrange to read: `ds=(v_Ave)(`dt) or, change in position equals average velocity times change in clock time. confidence rating #$&*: 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): ------------------------------------------------ Self-critique rating: #$&* ********************************************* Question: Given average speed and distance moved how do you find the corresponding time interval? YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: We rearrange again: v_Ave=(`ds)/(`dt) (v_Ave)(`dt)=(`ds) `dt=(`ds)/(v_Ave) change in clock time equals distance moved divided by average speed. confidence rating #$&*: 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): ------------------------------------------------ Self-critique rating: #$&* ********************************************* Question: Given time interval and distance moved how do you get average speed? YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: Average speed equals change in position divided by change in clock time. vAve=(`ds)/(`dt) confidence rating #$&*: ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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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): ------------------------------------------------ Self-critique rating: #$&* ********************************************* Question: A ball rolls from rest down a book, off that book and onto another book, where it picks up speed before rolling off the end of that book. Consider the interval that begins when the ball first encounters the second book, and ends when it rolls of the end of the book. For this interval, place in order the quantities initial velocity (which we denote v_0), and final velocity (which we denote v_f), average velocity (which we denote v_Ave). During this interval, the ball's velocity changes. It is possible for the change in its velocity to exceed the three quantities you just listed? Is it possible for all three of these quantities to exceed the change in the ball's velocity? Explain. Note that the change in the ball's velocity is denoted `dv. YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: lowest: initial velocity, v_0 middle: average velocity, v_Ave highest: final velocity, v_f The ball starts at a certain speed greater than zero and begins to increase in speed as soon as it starts rolling down the second book. The change in speed would be the difference between final and inital speed, or v_f minus v_0. Since v_f is the fastest speed during this interval (we know this because it is said to be picking up speed throughout the interval), and we are subtracting v_0 to get `dv, it must follow that `dv will be less than v_f. confidence rating #$&*: 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: v_Ave=(v_0+v_f)/2 =(4 m/s + 10 m/s)2 =7 m/s `dv=(v_f-v_0) =10 m/s - 4 m/s =6 m/s Speeds in order from least to greatest: v_0 = 4 m/s `dv = 6 m/s v_Ave = 7 m/s v_f = 10 m/s If v_0 = 9 m/s, then `dv = 1 m/s and v_Ave = .5 m/s. So the positions of average velocity and change in velocity would switch. It is not possible for the change in velocity to exceed the other three quantities. The change in velocity for two velocities greater than zero will always be less than the final velocity, as it is found by taking the difference between initial and final (ie subtracting a number greater than zero from the final velocity). confidence rating #$&*: 3 ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^ #$&* " Self-critique (if necessary): ------------------------------------------------ Self-critique rating: