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course Phy 241
003. Velocity Relationships*********************************************
Question: `q001. Note that there are 13 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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Your solution: meters/ sec
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Question: `q002. If the 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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Your solution: 'ds is measured in cm
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Question: `q003. Explain the algebra of multiplying the unit cm / sec by the unit sec.
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Your solution:
cm/sec * sec/1
You are trying to get ride of the unit sec in this problem. If you have x/x, it will = 1. If you multiply through our equation you will get
cm*sec / sec*1 you have sec/ sec * cm / 1 if you break it up in another way. Sec/sec = 1 and cm/1 = cm.
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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?
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Your solution: `dt = `ds / vAve. `dt = km / km/sec; `dt = seconds
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Question: `q005. Explain the algebra of dividing the unit km / sec into the unit km.
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Your solution:
km / km/sec can be broken up to look like km/1 * 1/(km/sec). to solve we can set it = to zero.
km/1 * 1/(km/sec) = 0
multiply both sides by (km/sec)/ 1: km = km/sec
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If you multiply both sides by km/s / 1 you get km = 0.
The unit is not zero and shouldn't be set equal to zero.
1 / (km/sec) = sec / km. Multiply this by km and you get sec.
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multiply each side by sec to create km/1 on the right side: sec*km = km
divide by km to get all like terms on one side: sec = km/km
km/km = 1 so we are left with seconds as our unit.
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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?
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Your solution: 4m at 2s and 10m at 5s
10-4 = 6m
5-2 = 3s
6m / 3s = 2m/s on average
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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.
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Your solution: `ds = s2 - s1 & `dt = t2 - t1 then `ds/`dt = vAve and (s2-s1)/(t2-t1) also equals vAve
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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?
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Your solution: (10m - 4m)/ (4s-2s) = 6m / 3s = 2m/s
The initial rise is of 6 meters. This means that the position changed by 6 meters. On a graph you would move ""up"" 6 units. The initial run is by
2 seconds. It took 2 seconds to move the 6 meters. On a graph, From the point you stopped at you would move to the right 2 units.
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Question: `q009. What is the slope of this triangle and what does it represent?
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Your solution: The slope is 2m/s (algebra in the above question). The slope states that for every run of 1 second you have a rise of 2 meters.
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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?
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Your solution: The greater the slope, the higher the rise and the shorter the run. The more position gained in the smaller amount of clock time
the greater the velocity.
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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?
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Your solution: The graph is increasing at an increasing rate. If slope increases then it states velocity is also increasing.
Question: `q012. If at clock time t = t_1 the position of an object is x = x_1, while at clock time t = t_2 its position is x = x_2, then what
is its average velocity during the corresponding interval?
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Your solution: `ds/ `dt = vAve = (x_2 - x_1) / (t_2 - t_1) = vAve
Question: `q013. On a graph of position x vs. clock time t, what is the average slope between the point (t_1, x_1) and (t_2, x_2)? What is
the meaning of this average slope, and why?
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Your solution: (x_2 - x_1) / (t_2 - t_1) = slope
slope is the change in clock time over the change in position."
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