#$&* course Phy 231 2/2 9:20pm 005. Uniformly Accelerated Motion
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Given Solution: The velocity of the object changes from 5 meters/second to 25 meters/second so the change in velocity is 20 meters/second. The average acceleration is therefore (20 meters/second) / (4 seconds) = 5 m / s^2. The average velocity of the object is the average of its initial and final velocities, as asserted above, and is therefore equal to (5 meters/second + 25 meters/second) / 2 = 15 meters/second (note that two numbers are averaged by adding them and dividing by 2). &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): ------------------------------------------------ Self-critique rating: ********************************************* Question: `q002. How far does the object of the preceding problem travel in the 4 seconds? YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: Average velocity is the rate of change in position with respect to time. vAve=`ds/`dt 15m/s=(`ds)/4s (15m/s)(4s)=`ds 60m=`ds The object travels 60m. confidence rating #$&*: ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: The displacement `ds of the object is the product vAve `dt of its average velocity and the time interval, so this object travels 15 m/s * 4 s = 60 meters during the 4-second time interval. &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): ------------------------------------------------ Self-critique rating: ********************************************* Question: `q003. Explain in commonsense terms how we determine the acceleration and distance traveled if we know the initial velocity v0, and final velocity vf and the time interval `dt. YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: We know that the object increases in speed from a velocity of v0 to a velocity of vf, and we assume that it is increasing at a consistent rate. So, we look at the time interval and see how much the velocity would have to change per unit of time within that interval in order to increase to vf in the given time. This means we need to divide the change in velocity, or (vf-v0), by the time interval. That will tell us the rate the velocity is changing per unit of time; that is the acceleration. To find the distance travelled, we first need to find the average speed during that interval by adding up the initial and final velocities and dividing by 2. This tells us the average distance per unit of time the object is moving. So now we can multiply by the number of units of time, which well tells us the distance the object travels over the entire interval. confidence rating #$&*: ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: In commonsense terms, we find the change in velocity since we know the initial and final velocities, and we know the time interval, so we can easily calculate the acceleration. Again since we know initial and final velocities we can easily calculate the average velocity, and since we know the time interval we can now determine the distance traveled. &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): ------------------------------------------------ Self-critique rating: ********************************************* Question: `q004. Symbolize the situation by first giving the expression for the acceleration in terms of v0, vf and `dt, then by giving the expression for vAve in terms of v0 and vf, and finally by giving the expression for the displacement in terms of v0, vf and `dt. YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: If given v0 and vf, we find their difference to calculate `dv. Now we can evaluate aAve=`dv/`dt. We find the sum of v0 and vf and divide by 2 to find vAve. vAve=(v0+vf)/2 Finally, we know that vAve=`ds/`dt, which can be rearranged as (vAve)(`dt)=`ds. Therefore, `ds=[(v0+vf)/2)](`dt) confidence rating #$&*: ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: The acceleration is equal to the change in velocity divided by the time interval; since the change in velocity is vf - v0 we see that the acceleration is a = ( vf - v0 ) / `dt. The average velocity is the average of the initial and final velocities, which is expressed as (vf + v0) / 2. When this average velocity is multiplied by `dt we get the displacement, which is `ds = (v0 + vf) / 2 * `dt. STUDENT SOLUTION (mostly but not completely correct) vAve = (vf + v0) / 2 aAve = (vf-v0) / dt displacement = (vf + v0)/dt INSTRUCTOR RESPONSE Displacement = (vf + v0)/dt is clearly not correct, since greater `dt implies greater displacement. Dividing by `dt would give you a smaller result for larger `dt. From the definition vAve = `ds / `dt, so the displacement must be `ds = vAve * `dt. Using your correct expression for vAve you get the correct expression for `ds. &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): ------------------------------------------------ Self-critique rating: ********************************************* Question: `q006. This situation is identical to the previous, and the conditions implied by uniformly accelerated motion are repeated here for your review: If the acceleration of an object is uniform, then the following statements apply: 1. A graph of velocity vs. clock time forms a straight line, either level or increasing at a constant rate or decreasing at a constant rate. 2. The average velocity of the object over any time interval is equal to the average of its velocity at the beginning of the time interval (called its initial velocity) and its velocity at the end of the time interval (called its final velocity). 3. The velocity of the object changes at a constant rate (this third statement being obvious since the rate at which the velocity changes is the acceleration, which is assumed here to be constant). 4. The acceleration of the object at every instant is equal to the average acceleration of the object. Describe a graph of velocity vs. clock time, assuming that the initial velocity occurs at clock time t = 0. At what clock time is the final velocity then attained? What are the coordinates of the point on the graph corresponding to the initial velocity (hint: the t coordinate is 0, as specified here; what is the v coordinate at this clock time? i.e., what is the velocity when t = 0?). What are the coordinates of the point corresponding to the final velocity? YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: Since the time interval of 4 seconds begins at t=0, the interval would end at t=4. So the clock time for final velocity would be t=4. The coordinate corresponding to initial velocity would occur at t=0, initial clock time. Initial velocity is given as 5m/s, so the coordinate would occur at point (0,5), with time on the x-axis and velocity on the y-axis. The coordinate corresponding to final velocity would occur at t=4, final clock time. Final velocity is given as 25m/s, so the ccoordinate would occur at point (4,25). confidence rating #$&*: ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: The initial velocity of 5 m/s occurs at t = 0 s so the corresponding graph point is (0 s, 5 m/s). The final velocity of 25 meters/second occurs after a time interval of `dt = 4 seconds; since the time interval began at t = 0 sec it ends at at t = 4 seconds and the corresponding graph point is ( 4 s, 25 m/s). &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): ??? In general, should we use units when naming coordinates on a graph ???
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Given Solution: Since the acceleration is uniform, the graph is a straight line. The graph therefore increases at a constant rate from the point (0, 5 m/s) to the point (4 s, 25 m/s). &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): ------------------------------------------------ Self-critique rating: ********************************************* Question: `q008. What is the slope of the graph between the two given points, and what is the meaning of this slope? YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: Suppose that an object increases its velocity at a uniform rate, from an initial velocity of 5 m/s to a final velocity of 25 m/s during a time interval of 4 seconds. Slope is calculated as rise/run. In this case, the rise is the change in velocity, which begins at 5m/s and increases to 25m/s for a rise of 20m/s. The run is the change in time, which begins at 0s and ends at 4s for a run of 4s. So, the rise/run is 20m/s/4s, or 5m/s^2. The slope of a velocity v. time graph shows the rate at which velocity is changing with respect to time, or the rate at which the object is accelerating. confidence rating #$&*: ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: The rise of the graph is from 5 m/s to 25 m/s and is therefore 20 meters/second, which represents the change in the velocity of the object. The run of the graph is from 0 seconds to 4 seconds, and is therefore 4 seconds, which represents the time interval during which the velocity changes. The slope of the graph is rise / run = ( 20 m/s ) / (4 s) = 5 m/s^2, which represents the change `dv in the velocity divided by the change `dt in the clock time and therefore represents the acceleration of the object. &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): ------------------------------------------------ Self-critique rating: ********************************************* Question: `q009. The graph forms a trapezoid, starting from the point (0,0), rising to the point (0,5 m/s), then sloping upward to (4 s, 25 m/s), then descending to the point (4 s, 0) and returning to the origin (0,0). This trapezoid has two altitudes, 5 m/s on the left and 25 m/s on the right, and a base which represents a width of 4 seconds. What is the average altitude of the trapezoid and what does it represent, and what is the area of the trapezoid and what does it represent? YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: The average altitude can be found by calculating the sum of the lower and higher altitudes and dividing by 2. In this case, the altitudes are initial velocity and final velocity, so our calculation here will lead us to the average velocity once again. (v0+vf)/2=(30m/s)2=15m/s. To find the area of the trapezoid, we multiply base by average height. Here, that's 4s*15m/s, the same as time interval * average velocity. We end up with a result of 60m, the total displacement of the object. This shows us that on a velocity v. time graph, the area under the graph equals the total displacement of the object. confidence rating #$&*: ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: The two altitudes are 5 meters/second and 25 meters/second, and their average is 15 meters/second. This represents the average velocity of the object on the time interval. The area of the trapezoid is equal to the product of the average altitude and the base, which is 15 m/s * 4 s = 60 meters. This represents the product of the average velocity and the time interval, which is the displacement during the time interval. STUDENT COMMENT I understand how to find the average altitude and multiply it by the amount of seconds. I also understand how to find the area of the trapezoid. But, again I don’t understand what it repreents, which is the product of the average velocity and the time interval, or the displacement. INSTRUCTOR RESPONSE If you multiply the average velocity on a time interval by the duration of the interval, you get the displacement. Since the average altitude represents the average velocity and the width represents the duration of the time interval, the product therefore represents the displacement. Since the product of average altitude and width is area, it follows that this product represents the displacement. &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): ------------------------------------------------ Self-critique rating: ********************************************* Question: `q010. Students at this point often need more practice identifying which of the quantities v0, vf, vAve, `dv, a, `ds and `dt are known in a situation or problem. You should consider running through the optional supplemental exercise ph1_qa_identifying_quantities.htm . The detailed URL is http://vhcc2.vhcc.edu/dsmith/genInfo/qa_query_etc/ph1/ph1_qa_identifying_quantities.htm If you are able to quickly identify all the quantities correctly 'in your head', the exercise won't take long and it won't be necessary to type in any responses or submit anything. If you aren't sure of some of the answers, you can submit the document, answer and/or asking questions on only the problems of which you are unsure. You should take a quick look at this document. Answer below by describing what you see and indicating whether or not you think you already understand how to identify the quantities. If you are not very sure you are able to do this reliably, indicate how you have noted this link for future reference. If you intend to submit all or part of the document, indicate this as well. YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: I feel good about my ability to solve these problems. But, I went about one of them completely differently. The formula you suggest for question 5, 'ds=v0`dt + .5a `dt^2, is totally unfamiliar to me. So I didn't feel good plugging into that formula, but I reasoned through using a combination of formulas (which I'm assuming this is). I came up with the right answer. Is it important that I know this formula? Or is knowing all the basic relationships ok?
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Given Solution: You should have responded in such a way that the instructor understands that you are aware of this document, have taken appropriate steps to note its potential usefulness, and know where to find it if you need it. &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): ------------------------------------------------ Self-critique rating: