course Phy 201 6/6 5 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 moved you set up the equation: average speed = change in distance / time. Then, you solve for the distance by multiplying the average speed and the time. It will give you the distance travelled. confidence rating #$&* 3 ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: ** You multiply average speed * time interval to find distance moved. For example, 50 miles / hour * 3 hours = 150 miles. ** Self-critique (if necessary): OK Question: Given average speed and distance moved how do you find the corresponding time interval? Your solution: Using the same equation from above, you would solve for time by dividing the distance travelled by the average speed. This will give you the time in whatever time interval you are using, seconds, etc. confidence rating #$&* 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? Your solution: To find the average speed you divide the change in distance by the time interval. confidence rating #$&*: 3 ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: ** Average speed = distance / change in clock time. This is the definition of average speed. For example if we travel 300 miles in 5 hours we have been traveling at an average speed of 300 miles / 5 hours = 60 miles / hour. ** &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): OK Question: A ball rolls from rest down a book, off that book and onto another book, where it picks up speed before rolling off the end of that book. Consider the interval that begins when the ball first encounters the second book, and ends when it rolls of the end of the book. For this interval, place in order the quantities initial velocity (which we denote v_0), and final velocity (which we denote v_f), average velocity (which we denote v_Ave). During this interval, the ball's velocity changes. It is possible for the change in its velocity to exceed the three quantities you just listed? Is it possible for all three of these quantities to exceed the change in the ball's velocity? Explain. Note that the change in the ball's velocity is denoted `dv. YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: Slowest: initial velocity Average velocity Fastest: final velocity It is possible for the acceleration to exceed the velocity. If you have a whole number change in velocity, like from 12 to 14 m/s in less than 1 second, then the acceleration would be greater than the velocities measured. In addition, it is also possible for the measured velocities to be greater than the acceleration, like if there was a change from 12 to 14 m/s in 15 seconds, the acceleration would be less. Whether these situations would actually happen is unknown and is based the slope of the incline and how the speed and intervals are measured. confidence rating #$&* 3 ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^ "