course PHY 121 7/22 003. Velocity Relationships*********************************************
.............................................
Given Solution: vAve = `ds / `dt. The units of `ds are cm and the units of `dt are sec, so the units of `ds / `dt must be cm / sec. Thus vAve is in cm/s. Self-critique (if necessary): OK Self-critique rating: ---- ********************************************* 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 must be measured in cm. Confidence rating: 3 ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
.............................................
Given Solution: Since vAve is in cm/sec and `dt in sec, `ds = vAve * `dt must be in units of cm / sec * sec = cm. &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): OK Self-critique rating: ---- ********************************************* Question: `q003. Explain the algebra of multiplying the unit cm / sec by the unit sec. YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: When multiplying cm/sec by sec, you can thing of the solitary “sec” as sec/1. If you multiply cm/sec * sec/1 across, you get cm*sec/sec. The sec’s cancel, leaving cm. Confidence rating: 3 ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
.............................................
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: ----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 sec. Confidence rating: 3 ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
.............................................
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: ---- ********************************************* Question: `q005. Explain the algebra of dividing the unit km / sec into the unit km. YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: We can think of km as km/1, and multiplying by the inverse is the same as dividing. So, multiplying across (km/1)*(sec/km) = (km*sec)/(km). Km divides out, and you’re left with sec. Confidence rating: 3 ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
.............................................
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): 3 Self-critique rating: ---- ********************************************* 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 `dt is 5-2=3 and the `ds = 10-4 =6. To get vAve, vAve = `ds/`dt = 3/6 = 2m/s. Confidence rating:3 ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
.............................................
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: ---- ********************************************* 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, when what expression represents the change `ds in position and what expression represents the change `dt in the clock time? YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: `ds = s2-s1, while `dt = t2-t1. Confidence rating: 2/3 ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
.............................................
Given Solution: 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. &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): I was headed in the right direction and think I got it but the last sentence of the given answer confuses me.
.............................................
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: ---- ********************************************* Question: `q009. What is the slope of this triangle and what does it represent? YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: slope = rise/run = 6/3 = 2m/s Confidence rating: 3 ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
.............................................
Given Solution: The slope of this graph is 6 meters / 3 seconds = 2 meters / second. &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): OK Self-critique rating: ---- ********************************************* 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 displays the relationship between the two values of position and time. Greater slope implies greater velocity because the a greater slope means that the denominator in the slope equation (`ds) is getting larger and larger compared to `dt. Confidence rating: 3 ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
.............................................
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: ---- ********************************************* 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: A graph of this car’s travel would be increasing at an increasing rate and would have a positive slope. Since the slope would be increasing upward, it reflects the fact that there is a greater distance developing between the values for `ds and `dt - - the car is increasing in speed. Confidence rating: 2/3 ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
.............................................
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): I didn’t describe it exactly that way but I understand. Self-critique rating: "