course Phy 121
`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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centimeters per second
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.
`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?
'ds must be measured in cm/sec*sec
Since vAve is in cm/sec and `dt in sec, `ds = vAve * `dt must be in units of cm / sec * sec = cm.
`q003. Explain the algebra of multiplying the unit cm / sec by the unit sec.
When we multiply cm/sec by sec we are eliminating the seconds which leaves the solution in centimeters.
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.
`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?
km / (km/sec)
Since `dt = `ds / vAve and `ds is in km and vAve in km/sec, `ds / vAve will be in km / (km / sec) = seconds.
`q005. Explain the algebra of dividing the unit km / sec into the unit km.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.
`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?
To obtain the average velocity we must divide the change in position by the change in time which would look like this: (10-4) / (5-2) = 6m/3sec = 2m/sec
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.
`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? 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.
vAve = (s2 - s1) / (t2 - t1)
`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?
The rise would equal the change in position which is 10m - 4m = 6m.
The run would equal the change in time which is 5s - 2s = 3s.
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.
`q009. What is the slope of this triangle and what does it represent?
2meters/second
The slope of this graph is 6 meters / 3 seconds = 2 meters / second.
`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?
Since the rise over run is equal to change in position over change in time the slope gives us average velocity.
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.
`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.
The graph would be increasing at an increasing rate.
Is the slope of your graph increasing or decreasing?
Increaseing
How does the behavior of the slope of your graph indicate the condition of the problem, namely that the velocity is increasing?
The slope is defining the increase in velocity of the car by showing the increase of change in position over time.
you should 'click through' this qa to see in detail how your responses compare with the given solutions. However it's clear that you understand this and you're doing very well.
I inserted this note before; I'm not hitting you over the head with the redundancy, just making sure you don't miss it:
&#To run the program correctly you click on the Next Question/Solution button (top left), enter your response in the Answer box (the box at top right), click on the Enter Response button (above the top right box), then again on the Next Question/Solution botton (top left) to see the solution, and finally enter your response or self-critique in the Answer box and click on Enter Response.
You then start the process over with the next question, clicking on the Next Question/Solution button at top left, etc. &#