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Given Solution: vAve = `ds / `dt. The units of `ds are m and the units of `dt are sec, so the units of `ds / `dt must be m / sec. Thus vAve is in m/s. &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): Ok ------------------------------------------------ Self-critique rating #$&* 3 ********************************************* Question: `q002. If the definition equation vAve = `ds / `dt is to be solved for `ds we multiply both sides of the equation by `dt to obtain `ds = vAve * `dt. If vAve is measured in cm / sec and `dt in sec, then in what units must `ds be measured? YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: 'ds, the change in displacement, must be measured in cm, if vAve is measured in cm/s and 'dt is measured in s. The seconds cancel out when multiplying both sides by 'dt to obtain 'ds. Verified: vAve = 'ds / 'dt vAve = x centimeters / y seconds vAve ( centimeters / second ) * 'dt ( seconds ) = 'ds ( centimeters ) confidence rating #$&* 3 ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: Since vAve is in cm/sec and `dt in sec, `ds = vAve * `dt must be in units of cm / sec * sec = cm. STUDENT QUESTION I don’t get how sec and sec would cancel each other out INSTRUCTOR RESPONSE cm / s * s means (cm/s) * s, which is the same as (cm / s) * (s / 1). Multiplying numerators and denominators we have (cm * s) / (s * 1) or just (cm * s) / s, which is the same as cm * (s / s) = cm * 1 = cm. &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): Ok ------------------------------------------------ Self-critique rating #$&* 3 ********************************************* Question: `q003. Explain the algebra of multiplying the unit cm / sec by the unit sec. YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: Visually, multiplying cm/sec. by sec. in a dimensional analysis style-like form, the sec. on the top crosses out with the sec. on the bottom. This occurs because of the concept of numerators and denominators in fractions. Anything over itself is always equal to 1. Any number or unit over itself simplifies to 1. This is concept is similar to inverse operations. With any number, if you multiply and then divide by the same number, the result is the original number before you did both operations. Thus, the terms are said to cancel out, or reduce themselves to 1. In this case, multiplying sec. by cm = cm * sec. Dividing by sec., the result is simply, cm. confidence rating #$&* 3 ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: 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. &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): Ok ------------------------------------------------ Self-critique rating #$&* 3 ********************************************* Question: `q004. If the definition vAve = `ds / `dt is to be solved for `dt we multiply both sides of the equation by `dt to obtain vAve * `dt = `ds, then divide both sides by vAve to get `dt = `ds / vAve. If vAve is measured in km / sec and `ds in km, then in what units must `dt be measured? YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: 'dt must be measured in seconds. The km units ""cancel out"" in 'ds and vAve, making the 'dt expressed in seconds. confidence rating #$&* 3 ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: Since `dt = `ds / vAve and `ds is in km and vAve in km/sec, `ds / vAve will be in km / (km / sec) = seconds. &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): Ok ------------------------------------------------ Self-critique rating #$&* 3 ********************************************* Question: `q005. Explain the algebra of dividing the unit km / sec into the unit km. YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: Dividing km by km/ sec., the km units cancel out, leaving seconds. This can be seen visually by using the reciprocity of multiplication and division. In order to divide km by km/sec., you multiply km over 1 by the reciprocal of km/sec., which is sec./km, which equates to seconds over 1, or simply seconds. confidence rating #$&* 3 ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: 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. &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): Ok ------------------------------------------------ Self-critique rating #$&* 3 ********************************************* Question: `q006. If an object moves from position s = 4 meters to position s = 10 meters between clock times t = 2 seconds and t = 5 seconds, then at what average rate is the position of the object changing (i.e., what is the average velocity of the object) during this time interval? What is the change `ds in position, what is the change `dt in clock time, and how do we combine these quantities to obtain the average velocity? YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: The change in position, 'ds, is 6 meters ( 10m - 4m ), or 6m. The change in clock time, 'dt, is 3 seconds ( 5s - 2s ), or 3 s. To find the average velocity, we use the formula vAve = 'ds / 'dt. vAve = 'ds / 'dt vAve = 6m / 3s vAve = 2m/ s The average velocity is 2 meters per second. confidence rating #$&* 3 ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: 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. &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): Ok ------------------------------------------------ Self-critique rating #$&* 3 ********************************************* Question: `q007. Symbolize this process: If an object moves from position s = s1 to position s = s2 between clock times t = t1 and t = t2, then what expression represents the change `ds in position and what expression represents the change `dt in the clock time? What expression therefore symbolizes the average velocity between the two clock times. YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: 'ds = s2 - s1 'dt = t2 - t1 vAve = 'ds / 'dt vAve = ( s2 - s1 ) / ( t2 - t1 ) confidence rating #$&* 3 ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: 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. The symbolic expression for the average velocity is therefore • vAve = `ds / `dt = (s2 - s1) / (t2 - t1). &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): Ok ------------------------------------------------ Self-critique rating #$&* 3 ********************************************* Question: `q008. On a graph of position s vs. clock time t we see that the first position s = 4 meters occurs at clock time t = 2 seconds, which corresponds to the point (2 sec, 4 meters) on the graph, while the second position s = 10 meters occurs at clock time t = 5 seconds and therefore corresponds to the point (5 sec, 10 meters). If a right triangle is drawn between these points on the graph, with the sides of the triangle parallel to the s and t axes, the rise of the triangle is the quantity represented by its vertical side and the run is the quantity represented by its horizontal side. This slope of the triangle is defined as the ratio rise / run. What is the rise of the triangle (i.e., the length of the vertical side) and what quantity does the rise represent? What is the run of the triangle and what does it represent? YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: The run of the triangle is 3 seconds ( 5 sec - 2 sec ); it shows an increase of the x values ( x2 - x1 ), clock time. The rise is 6 meters ( 10 m - 4 m ); it shows an increase of the y values ( y2 - y1 ), length. confidence rating #$&* 3 ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: 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. &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): Ok ------------------------------------------------ Self-critique rating #$&* 3 ********************************************* Question: `q009. What is the slope of this triangle and what does it represent? YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: The slope of the triangle, the ratio of rise to run, is 6/3 or 2 meters to every second; it represents by how much the length increases every second. confidence rating #$&* 3 ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: The slope of this graph is 6 meters / 3 seconds = 2 meters / second. &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): Ok ------------------------------------------------ Self-critique rating #$&* 3 ********************************************* Question: `q010. In what sense does the slope of any graph of position vs. clock time represent the velocity of the object? For example, why does a greater slope imply greater velocity? YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: The slope of a graph represents the velocity of an object because it illustrates the rate at which an object's position changes with time, the 'ds and 'dt, which is equal to rise over run. This is equal and derived from the average velocity. confidence rating #$&* 3 ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: Since the rise between two points on a graph of velocity vs. clock time represents the change `ds in position, and since the run represents the change `dt clock time, the slope represents rise / run, or change in position / change in clock time, which is `ds / `dt. This is equal to the average rate of change of position with respect to clock time, which is the definition of average velocity. &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): Ok ------------------------------------------------ Self-critique rating #$&* 3 ********************************************* Question: `q011. As a car rolls from rest down a hill, its velocity increases. Describe a graph of the position of the car vs. clock time. If you have not already done so, tell whether the graph is increasing at an increasing rate, increasing at a decreasing rate, decreasing at an increasing rate, decreasing at a decreasing rate, increasing at a constant rate or decreasing at a constant rate. Is the slope of your graph increasing or decreasing? How does the behavior of the slope of your graph indicate the condition of the problem, namely that the velocity is increasing? YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: The graph would be positive and its slope would be increasing, at an increasing rate, because the car is continually moving faster and faster and farther and farther away from the top of the hill, its position of rest. confidence rating #$&* 3 ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: 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 (an alternative description would be that the graph is increasing and concave up). &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): Ok ------------------------------------------------ Self-critique rating #$&* 3 You should submit the above questions, along with your answers and self-critiques. You may then begin work on the Questions, Problem and Exercises, as instructed below. "