#$&* course Phy 121 Question: `q001. Note that there are 13 questions in this assignment.
.............................................
Given Solution: The velocity of the object changes from 5 meters/second to 25 meters/second so the change in velocity is 20 meters/second. The average acceleration is therefore (20 meters/second) / (4 seconds) = 5 m / s^2. The average velocity of the object is the average of its initial and final velocities, as asserted above, and is therefore equal to (5 meters/second + 25 meters/second) / 2 = 15 meters/second (note that two numbers are averaged by adding them and dividing by 2). &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): ok ------------------------------------------------ Self-critique rating: ok ********************************************* Question: `q002. How far does the object of the preceding problem travel in the 4 seconds? YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: In 4 seconds, the equation would look like this: 15m/s * 4s/1 = 60 meters. confidence rating #$&*: ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
.............................................
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): ok ------------------------------------------------ Self-critique rating: ok ********************************************* 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 I have a 5.56mm round loaded into an M16A2 service rifle, with an initial velocity of zero and an initial time interval of zero. When I pull the trigger, in a matter of 0.000486 seconds the round has a final muzzle velocity of 3600 feet per second. So I have two points of data, with an initial velocity and a final muzzle velocity which means that I know the displacement of the round in the barrel and the amount of time that it took to get from the shell casing to the tip of the muzzle. The information needed to find acceleration is two different points of interest that can each provide a velocity and a clock time at which that velocity occurred. I would divide the difference in the velocity 2 and velocity 1(respectively) by the difference in clock time 2 and clock time one(again, respectively) to find the acceleration of the round in meters per second^2 with reference to the two points. If we know the initial and final velocities and the change in time between the two, then we can calculate the distance traveled by multiplying the average velocity by the time interval using meters per second^2 as the unit again. confidence rating #$&*: ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
.............................................
Given Solution: In commonsense terms, we find the change in velocity since we know the initial and final velocities, and we know the time interval, so we can easily calculate the acceleration. Again since we know initial and final velocities we can easily calculate the average velocity, and since we know the time interval we can now determine the distance traveled. &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): I guess a bit overboard in my explanation. Funny how those old Marine Corps riffle range figures pop up sometimes. ------------------------------------------------ Self-critique rating: Ok ********************************************* Question: `q004. Symbolize the situation by first giving the expression for the acceleration in terms of v0, vf and `dt, then by giving the expression for vAve in terms of v0 and vf, and finally by giving the expression for the displacement in terms of v0, vf and `dt. YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: acceleration = vf - v0 / `dt vAve = vf + v0 / 2 `ds = (vf + v0) / 2 * `dt confidence rating #$&*: ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
.............................................
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): ok ------------------------------------------------ Self-critique rating: ok ********************************************* 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: The velocity would be represented on the y axis, and the time on the x axis. The initial velocity would then be (0s,5m/s) and the final velocity would then be (4s, 25m/s). So the final velocity is attained at 4 seconds. confidence rating #$&*: ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
.............................................
Given Solution: The initial velocity of 5 m/s occurs at t = 0 s so the corresponding graph point is (0 s, 5 m/s). The final velocity of 25 meters/second occurs after a time interval of `dt = 4 seconds; since the time interval began at t = 0 sec it ends at at t = 4 seconds and the corresponding graph point is ( 4 s, 25 m/s). &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): ok ------------------------------------------------ Self-critique rating: ok ********************************************* Question: `q007. Is the v vs. t graph increasing, decreasing or level between the two points, and if increasing or decreasing is the increase or decrease at a constant, increasing or decreasing rate? YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: The graph is increasing at a uniform rate, so the graph of the line would be increasing at a constant rate. confidence rating #$&*: ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
.............................................
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: ok ********************************************* 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 of the graph equals (25m/s - 5m/s) / (4s - 0s) = 20m/s / 4s = 5m/s^2. It means that the object is increasing in velocity at a rate of 5m/s for each second(change in x). The slope tells us the acceleration of the object. confidence rating #$&*: ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
.............................................
Given Solution: The rise of the graph is from 5 m/s to 25 m/s and is therefore 20 meters/second, which represents the change in the velocity of the object. The run of the graph is from 0 seconds to 4 seconds, and is therefore 4 seconds, which represents the time interval during which the velocity changes. The slope of the graph is rise / run = ( 20 m/s ) / (4 s) = 5 m/s^2, which represents the change `dv in the velocity divided by the change `dt in the clock time and therefore represents the acceleration of the object. &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): ok ------------------------------------------------ Self-critique rating: ok ********************************************* Question: `q009. The graph forms a trapezoid, starting from the point (0,0), rising to the point (0,5 m/s), then sloping upward to (4 s, 25 m/s), then descending to the point (4 s, 0) and returning to the origin (0,0). This trapezoid has two altitudes, 5 m/s on the left and 25 m/s on the right, and a base which represents a width of 4 seconds. What is the average altitude of the trapezoid and what does it represent, and what is the area of the trapezoid and what does it represent? YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: The average altitude is 25 + 5 = 30, 30 / 2 = 15. The average altitude represents the vAve. The area under the trapezoid is Area = base * height, so Area = 4s * 15m/s = 60m. The area(being in meters) gives us the change in position during the change in time. confidence rating #$&*: ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
.............................................
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: ok ********************************************* 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: It was a query(looked like one I'll be doing soon). I was able to pick out the data. I bookmarked it as id quantities just in case. confidence rating #$&*: ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
.............................................
Given Solution: You should have responded in such a way that the instructor understands that you are aware of this document, have taken appropriate steps to note its potential usefulness, and know where to find it if you need it. &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): ------------------------------------------------ Self-critique rating: ********************************************* Question: `q011. The velocity of a car changes uniformly from 5 m/s to 25 m/s during an interval that lasts 6 seconds. Show in detail how to reason out how far it travels. YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: If I know the cars average velocity over a given time, then I can tell you how far its moved because velocity is change in position per unit change in time. So, if the cars average velocity is (5m/s + 25m/s) / 2 = 15m/s and it did so over a 6 second time frame then; 15m/s * 6s = 90meters traveled. confidence rating #$&*: ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^ ********************************************* 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 average graph altitude is (10m/s + 20m/s) / 2 = 15m/s. The average graph altitude is the average velocity of the graph with relationship to the change in time. It has this meaning because it is the point that rests between the two points on the curve of the graph. The area of the trapezoid is the area under the graph between the time intervals. It gives us the distance traveled by multiplying the base of the trapezoid(tf - t0) by the height of the trapezoid(the average graph altitude or the average velocity). The average velocity has units of m/s which would be multiplied by seconds(change in time), which would leave us with meters(length). confidence rating #$&*: ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^ ********************************************* Question: `q013. On a certain interval of duration `dt an object has initial velocity v_0 and final velocity v_f. In terms of the symbols v_0, v_f and `dt, what are the values of the following? vAve `dv `ds aAve Be sure to explain your reasoning. YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: vAve = (v_0 + v_f) / 2, The average velocity is the altitude average of the graph. `dv = v_f - v_0, The change in velocity is equal to the final velocity minus the initial velocity. `ds = vAve * `dt, The change in position is equal to the average velocity multiplied by the change in time(time interval). aAve = `dv / `dt, The average acceleration is equal to the change in velocity divided by the change in time. " Self-critique (if necessary): ------------------------------------------------ Self-critique rating: ********************************************* Question: `q011. The velocity of a car changes uniformly from 5 m/s to 25 m/s during an interval that lasts 6 seconds. Show in detail how to reason out how far it travels. YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: If I know the cars average velocity over a given time, then I can tell you how far its moved because velocity is change in position per unit change in time. So, if the cars average velocity is (5m/s + 25m/s) / 2 = 15m/s and it did so over a 6 second time frame then; 15m/s * 6s = 90meters traveled. confidence rating #$&*: ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^ ********************************************* 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 average graph altitude is (10m/s + 20m/s) / 2 = 15m/s. The average graph altitude is the average velocity of the graph with relationship to the change in time. It has this meaning because it is the point that rests between the two points on the curve of the graph. The area of the trapezoid is the area under the graph between the time intervals. It gives us the distance traveled by multiplying the base of the trapezoid(tf - t0) by the height of the trapezoid(the average graph altitude or the average velocity). The average velocity has units of m/s which would be multiplied by seconds(change in time), which would leave us with meters(length). 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 = (v_0 + v_f) / 2, The average velocity is the altitude average of the graph. `dv = v_f - v_0, The change in velocity is equal to the final velocity minus the initial velocity. `ds = vAve * `dt, The change in position is equal to the average velocity multiplied by the change in time(time interval). aAve = `dv / `dt, The average acceleration is equal to the change in velocity divided by the change in time. " Self-critique (if necessary): ------------------------------------------------ Self-critique rating: #*&!