course Phy 121 wEϚҝwӄassignment #003
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04:40:20
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RESPONSE --> vAve = 'ds 'dt vAve = 10 -4 5-2 (10 & 4 meters, 2 & 5 secs.) vAve = 6 3 vAve = 2 meters per second change in 'ds = +6 meters change in 'dt = +3 seconds
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04:41:35 `q001. Note that there are 11 questions in this assignment. vAve = `ds / `dt, which is the definition of average velocity and which fits well with our intuition about this concept. If displacement `ds is measured in meters and the time interval `dt is measured in seconds, in what units will vAve be obtained?
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RESPONSE --> Average velocity is obtained by dividing the displacement of an object by the amount of time that passed (vAve = 'ds/dt), so the units will be measured in terms of meters per second. confidence assessment: 3
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04:42:36 `q002. If the definition equation vAve = `ds / `dt is to be solved for `ds we multiply both sides of the equation by `dt to obtain `ds = vAve * `dt. If vAve is measured in cm / sec and `dt in sec, then in what units must `ds be measured?
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RESPONSE --> 'ds would be measured in centimeters confidence assessment: 3
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04:42:43 Since vAve is in cm/sec and `dt in sec, `ds = vAve * `dt must be in units of cm / sec * sec = cm.
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RESPONSE --> ok self critique assessment: 2
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04:44:00 `q003. Explain the algebra of multiplying the unit cm / sec by the unit sec.
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RESPONSE --> This process would involve cross-multiplication, which, in this case, would show as: cm sec ------ ------- sec 1 confidence assessment: 2
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04:44:29 When we multiply cm/sec by sec we are multiplying the fractions cm / sec and sec / 1. When we multiply fractions we will multiply numerators and denominators. We obtain cm * sec / ( sec * 1). This can be rearranged as (sec / sec) * (cm / 1), which is the same as 1 * cm / 1. Since multiplication or division by 1 doesn't change a quantity, this is just equal to cm.
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RESPONSE --> ok self critique assessment: 2
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04:45:05 `q004. If the definition vAve = `ds / `dt is to be solved for `dt we multiply both sides of the equation by `dt to obtain vAve * `dt = `ds, then divide both sides by vAve to get `dt = `ds / vAve. If vAve is measured in km / sec and `ds in km, then in what units must `dt be measured?
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RESPONSE --> 'dt would be measured in seconds confidence assessment: 3
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04:45:16 Since `dt = `ds / vAve and `ds is in km and vAve in km/sec, `ds / vAve will be in km / (km / sec) = seconds.
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RESPONSE --> ok self critique assessment: 2
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04:45:55 `q005. Explain the algebra of dividing the unit km / sec into the unit km.
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RESPONSE --> It sounds like the same principle of cross-multiplying: km sec ------ ------- sec 1 confidence assessment: 3
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04:46:35 The division is km / (km / sec). Since division by a fraction is multiplication by the reciprocal of the fraction, we have km * (sec / km). This is equivalent to multiplication of fractions (km / 1) * (sec / km). Multiplying numerators and denominators we get (km * sec) / (1 * km), which can be rearranged to give us (km / km) * (sec / 1), or 1 * sec / 1, or just sec.
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RESPONSE --> Looks similar to the answer given for the third question self critique assessment: 2
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04:48:34 `q006. If an object moves from position s = 4 meters to position s = 10 meters between clock times t = 2 seconds and t = 5 seconds, then at what average rate is the position of the object changing (i.e., what is the average velocity of the object) during this time interval? What is the change `ds in position, what is the change `dt in clock time, and how do we combine these quantities to obtain the average velocity?
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RESPONSE --> vAve = 'ds/'dt vAve = 10-4 [meters] / 5-2 [seconds] vAve = 6/3 vAve= 2 meters per second Change in 'ds is 6 meters, and the change in 'dt is a rate of 3 seconds confidence assessment: 3
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04:49:01 We see that the changes in position and clock time our `ds = 10 meters - 4 meters = 6 meters and `dt = 5 seconds - 2 seconds = 3 seconds. We see also that the average velocity is vAve = `ds / `dt = 6 meters / (3 seconds) = 2 meters / second. Comment on any discrepancy between this reasoning and your reasoning.
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RESPONSE --> Looks like the same answer I obtained :) self critique assessment: 2
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04:51:12 `q007. Symbolize this process: If an object moves from position s = s1 to position s = s2 between clock times t = t1 and t = t2, when what expression represents the change `ds in position and what expression represents the change `dt in the clock time?
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RESPONSE --> The only thing I can think of right now would be this: 'ds = s2-s1 'dt = t2-t1 confidence assessment: 2
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04:51:46 We see that the change in position is `ds = s2 - s1, obtained as usual by subtracting the first position from the second. Similarly the change in clock time is `dt = t2 - t1. What expression therefore symbolizes the average velocity between the two clock times.
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RESPONSE --> vAve = s2-s1/t2-t1 self critique assessment: 3
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05:02:53 `q008. On a graph of position s vs. clock time t we see that the first position s = 4 meters occurs at clock time t = 2 seconds, which corresponds to the point (2 sec, 4 meters) on the graph, while the second position s = 10 meters occurs at clock time t = 5 seconds and therefore corresponds to the point (5 sec, 10 meters). If a right triangle is drawn between these points on the graph, with the sides of the triangle parallel to the s and t axes, the rise of the triangle is the quantity represented by its vertical side and the run is the quantity represented by its horizontal side. This slope of the triangle is defined as the ratio rise / run. What is the rise of the triangle (i.e., the length of the vertical side) and what quantity does the rise represent? What is the run of the triangle and what does it represent?
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RESPONSE --> The equation for determining the slope on a graph is: m = y2--y1 x2-x1 [with y representing meters, and x being seconds] m = 10 -4 5 -2 m = 6 3 The rise of the triangle would be 6 meters (found by y2-y1] and is the opposite length of the triangle. The run would be 3 seconds (x2-x1) and it is the distance traveled (is also adjacent side of the triangle) confidence assessment: 3
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05:03:37 The rise of the triangle represents the change in the position coordinate, which from the first point to the second is 10 m - 4 m = 6 m. The run of the triangle represents the change in the clock time coordinate, which is 5 s - 2 s = 3 s.
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RESPONSE --> I see...guess I wasn't thinking in terms of a coordinate change...I need to remember that self critique assessment: 2
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05:04:58 `q009. What is the slope of this triangle and what does it represent?
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RESPONSE --> m = 10-46-3 m = 2 Slope measures the steepness of the change of line and coordinates confidence assessment: 1
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05:05:23 The slope of this graph is 6 meters / 3 seconds = 2 meters / second.
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RESPONSE --> ok self critique assessment: 2
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05:06:40 `q010. In what sense does the slope of any graph of position vs. clock time represent the velocity of the object? For example, why does a greater slope imply greater velocity?
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RESPONSE --> A greater slope means that the velocity of an object is changing at a faster rate per sec/min [or per whatever time frame is being used] confidence assessment: 2
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05:07:14 Since the rise between two points on a graph of velocity vs. clock time represents the change in `ds position, and since the run represents the change `dt clock time, the slope represents rise / run, or change in position /change in clock time, or `ds / `dt. This is the definition of average velocity.
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RESPONSE --> OK...just goes back to the vAve = 'ds/'dt equation self critique assessment: 2
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05:09:01 `q011. As a car rolls from rest down a hill, its velocity increases. Describe a graph of the position of the car vs. clock time. If you have not already done so, tell whether the graph is increasing at an increasing rate, increasing at a decreasing rate, decreasing at an increasing rate, decreasing at a decreasing rate, increasing at a constant rate or decreasing at a constant rate. Is the slope of your graph increasing or decreasing? How does the behavior of the slope of your graph indicate the condition of the problem, namely that the velocity is increasing?
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RESPONSE --> The slope of the graph would be increasing. The measure of the slope starts out gradually, then increases because the force of gravity (per rolling down the hill) causes velocity to increase confidence assessment: 2
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05:09:24 The graph should have been increasing, since the position of the car increases with time (the car gets further and further from its starting point). The slope of the graph should have been increasing, since it is the slope of the graph that indicates velocity. An increasing graph within increasing slope is said to be increasing at an increasing rate.
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RESPONSE --> Makes sense to me :) self critique assessment: 2
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