#$&* course PHY 121 6/15 11 005. Uniformly Accelerated Motion
.............................................
Given Solution: The velocity of the object changes from 5 meters/second to 25 meters/second so the change in velocity is 20 meters/second. The average acceleration is therefore (20 meters/second) / (4 seconds) = 5 m / s^2. The average velocity of the object is the average of its initial and final velocities, as asserted above, and is therefore equal to (5 meters/second + 25 meters/second) / 2 = 15 meters/second (note that two numbers are averaged by adding them and dividing by 2). &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): OK ------------------------------------------------ Self-critique rating: 3 ********************************************* Question: `q002. How far does the object of the preceding problem travel in the 4 seconds? YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: Every second it is traveling 5 m/s, so for 4 seconds, it would be 5 m/s * 4 s = 20 m. confidence rating #$&*: ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
.............................................
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): I knew that I was off when I had the m/s^2, but didn't stop to think it through. To find out how far it traveled, I needed the average speed times the seconds not the acceleration times the seconds. ------------------------------------------------ Self-critique rating: 3 ********************************************* 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: If you have a certain interval of time where you know how fast an object is moving at the beginning and how fast the object is moving at the end, you can find the amount of distance traveled by averaging those 2 speeds and multiplying by the amount of time that has elapsed. To find the acceleration, you would take the first velocity and subtract it from the second speed and then divide that difference by the amount of time that elapsed. confidence rating #$&*: ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
.............................................
Given Solution: In commonsense terms, we find the change in velocity since we know the initial and final velocities, and we know the time interval, so we can easily calculate the acceleration. Again since we know initial and final velocities we can easily calculate the average velocity, and since we know the time interval we can now determine the distance traveled. &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): OK ------------------------------------------------ Self-critique rating: 3 ********************************************* Question: `q004. Symbolize the situation by first giving the expression for the acceleration in terms of v0, vf and `dt, then by giving the expression for vAve in terms of v0 and vf, and finally by giving the expression for the displacement in terms of v0, vf and `dt. YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: aAve = (vf-v0)/'dt vAve = (v0+vf)/2 'ds=(vf-v0) * 'dt confidence rating #$&*: ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
.............................................
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): On the final one, I forgot to divide by 2 to get the average before I multiplied by the time. I do better when I use actual numbers at this point. I think if I'd done that with numbers, I would have said, ""That answer can't be right."" I'm still working on these formulas and how they relate.
.............................................
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): I guess I should also put the units of measure in the ordered pairs. I did not think that was usually done. ------------------------------------------------ Self-critique rating: 3
.............................................
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): OK ------------------------------------------------ Self-critique rating: 3 ********************************************* 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: The slope is 4 m/s / 1 s. You get this from the rise of 25 m/s - 5 m/s = 20 m/s, and the run of 4 s - 0 s = 4 s. 20 m/s / 4 s = 5 m/s^2 confidence rating #$&*: ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
.............................................
Given Solution: The rise of the graph is from 5 m/s to 25 m/s and is therefore 20 meters/second, which represents the change in the velocity of the object. The run of the graph is from 0 seconds to 4 seconds, and is therefore 4 seconds, which represents the time interval during which the velocity changes. The slope of the graph is rise / run = ( 20 m/s ) / (4 s) = 5 m/s^2, which represents the change `dv in the velocity divided by the change `dt in the clock time and therefore represents the acceleration of the object. &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): OK ------------------------------------------------ Self-critique rating: 3 ********************************************* Question: `q009. The graph forms a trapezoid, starting from the point (0,0), rising to the point (0,5 m/s), then sloping upward to (4 s, 25 m/s), then descending to the point (4 s, 0) and returning to the origin (0,0). This trapezoid has two altitudes, 5 m/s on the left and 25 m/s on the right, and a base which represents a width of 4 seconds. What is the average altitude of the trapezoid and what does it represent, and what is the area of the trapezoid and what does it represent? YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: The average altitude of the trapezoid (5 m/s + 25 m/s)/2 = 30 m/s / 2 = 15 m/s. This represents the average velocity of the object during this time interval. The area of the trapezoid is base times average height or 15 m/s * 4 s = 60 m, which is the amount of displacement of the object during this 4 second time frame. confidence rating #$&*: ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
.............................................
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. &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): OK ------------------------------------------------ Self-critique rating: 3 ********************************************* 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 was able to access the first link but not the second. I went through and wrote out what each formula meant so that I could associate words with the symbols. I substituted the information that was given into the formula to solve. At first, I just substituted the numbers and then worked from there without rearranging the formulas. After the first few I tried rearranging the formulas first before I plugged in the numbers. I generally got a good start and then had some minor difficulty with my math or I was on the right track but lacked self confidence in my solution. I copied and pasted this document onto a Word Document so that I can refer back to it. I did all of my work on the paper and marked on the ""Given Solution"" where I was on track and where I had difficulty. confidence rating #$&*: ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
.............................................
Given Solution: You should have responded in such a way that the instructor understands that you are aware of this document, have taken appropriate steps to note its potential usefulness, and know where to find it if you need it. &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): OK ------------------------------------------------ Self-critique rating: 2 If you understand the assignment and were able to solve the previously given problems from your worksheets, you should be able to complete most of the following problems quickly and easily. If you experience difficulty with some of these problems, you will be given notes and we will work to resolve difficulties. ********************************************* Question: `q011. The velocity of a car changes uniformly from 5 m/s to 25 m/s during an interval that lasts 6 seconds. Show in detail how to reason out how far it travels. YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: First you have to find the average velocity of the car. You do this by averaging the initial and final velocities. 5 m/s + 25 m/s = 30 m/s 30 m/s / 2 = 15 m/s. Then you have to multiply the average velocity by the elapsed time. 15 m/s * 6 s = 90 meters. confidence rating #$&*: ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^ ********************************************* Question: `q012. The points (5 s, 10 m/s) and (10 s, 20 m/s) define a 'graph trapezoid' on a graph of velocity vs. clock time. What is the average 'graph altitude' for this trapezoid? Explain what the average 'graph altitude' means and why it has this meaning. What is the area of this trapezoid? Explain thoroughly how you reason out this result, and be sure to include and explain your units. YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: The graph altitude is the average of the two vertical sides of the trapezoid. 10 m/s + 20 m/s = 30 m/s; 30 m/s / 2 = 15 m/s. This represents the average velocity of the object during this time interval. You obtain the average for the sides of the trapezoid in the same way you would mathematically obtain the average velocity by taking the initial velocity and the final velocity, adding them and dividing that sum by 2. In order to find the area, you have to multiply the average height that you just found by the elapsed time. In this case the elapsed time is 10 s - 5 s = 5 s. When you multiply 15 m/s * 5 s, you will get 75 meters. This represents the distance the object moved during these 5 seconds at this average velocity. confidence rating #$&*: ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^ ********************************************* Question: `q013. On a certain interval of duration `dt an object has initial velocity v_0 and final velocity v_f. In terms of the symbols v_0, v_f and `dt, what are the values of the following? vAve `dv `ds aAve Be sure to explain your reasoning. YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: vAve means average velocity. To find average velocity, you add the initial velocity (v_0) to the final velocity (v_f) and divide the sum by 2 vAve=(v_0 + v_f)/2 'dv means the change in velocity. To find the change in velocity, you subtract the initial velocity (v_0) from the final velocity (v_f) 'dv=v_f-v_0 'ds means change in distance. To find the change in distance, you find the average of the velocity (above) and multiply by the elapsed time ('dt) 'ds=(v_0+v_f)/2 * 'dt aAve means the average acceleration. To find the average acceleration, you find the change in velocity (above) and divide by the elapsed time ('dt) aAve=(v_f-v_0)/'dt confidence rating #$&*: ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^ Self-critique (if necessary): ------------------------------------------------ Self-critique rating: