course Phy 121 Had some trouble with this one. I am going back to study more. 鎦~EQecassignment #005
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20:45:37 `q001. Note that there are 9 questions in this assignment. 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. 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. By how much does the velocity of the object change? What is the average acceleration of the object? What is the average velocity of the object?
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RESPONSE --> The velocity changes by an increase of 20 m/s. The aAve of the object would be calculated as follows: aAve = (25-5)/4 = 5 m/s The vAve of the object would be calculated as follows: (25-5)/4 = 5 m/s confidence assessment: 3
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20:49:10 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.
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RESPONSE --> Since the average velocity of the object is 15 m/s we would multiply this number by 4 seconds and that would be 60 meters traveled in 4 seconds. self critique assessment: 3
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21:05:39 `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.
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RESPONSE --> Say my car was traveling 5 m/s and I speed up to 25 m/s and want to know how far I have traveled. Since 5 m/s is my v0 or initial velocity and 25 m/s is my vf or final velocity we know our 'ds or change in position which with a certain amount of time we can find a certain distance traveled. confidence assessment: 3
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21:11:15 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.
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RESPONSE --> Acceleration is equal to change in velocity ('dv) divided by a certain time interval and since Final velocity minus Initial velocity gives us change in velocity we find that we can solve acceleration by dividing final velocity minus initial velocty by change in time. self critique assessment: 3
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21:19:00 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).
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RESPONSE --> Since our initial clock time is 0 our coordinate for this position on our graph is (0, 5) and since our final velocity occurred 4 seconds later our final velocity would occur at clock time 4 seconds and our coordinate on the graph would be (4, 25). self critique assessment: 3
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21:19:34 `q007. Is the 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?
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RESPONSE --> The graph is increasing at a constant rate. confidence assessment: 3
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21:24:31 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.
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RESPONSE --> Since slope is equal to rise divided by run our slope would be equal to (25-5)/(4-0) which equals 5 m/s^2. This tells us that the object is increasing, which is its acceleration. self critique assessment: 3
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21:29:13 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.
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RESPONSE --> Since one side is 5 and the other is 25 we would add the two and divide by 2 which equals 15 m/s. I am not sure exactly what this means other than its velocity. Since the area of a trapezoid is equal to the average altitude multiplied by the base we get 60 meters. I am not sure what this means either other than maybe time. self critique assessment: 2
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twyٕݐw assignment #006 006. Using equations with uniformly accelerated motion. Physics I 02-03-2008
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22:16:02 The equation vf = v0 + a * `dt is solved for a by first adding -v0 to both sides to obtain vf - v0 = v0 + a * `dt - v0, which simplifies to vf - v0 = a * `dt. Both sides are then divided by `dt to obtain (vf - v0) / `dt = a. Reversing left-and right-hand sides we obtain the formula a = (vf - v0) / `dt. We then plug in our given values of initial and final velocities and the time interval. Since velocity increases from 10 m/s to 30 m/s, initial velocity is v0 = 10 m/s and final velocity is vf = 30 m/s. The time interval `dt is 15 seconds, so we have a = (30 m/s - 10 m/s) / (15 s) = 20 m/s / (15 s) = 1.33.. m/s^2.
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RESPONSE --> a = (vf-v0)/'dt = (30-20)/15 = 10/15 or 2/3 m/s^2 self critique assessment: 3
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22:17:10 `q002. It wasn't necessary to use a equation to solve this problem. How could this problem had been reasoned out without the use of an equation?
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RESPONSE --> You would just deivide the change in velocity by the time. confidence assessment: 2
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22:22:24 `q003. Use the equation `ds = (vf + v0) / 2 * `dt to determine the initial velocity of an object which accelerates uniformly through a distance of 80 meters in 10 seconds, ending up at a velocity of 6 meters / sec. begin by solving the equation for the desired quantity. Show every step of your solution.
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RESPONSE --> 'ds = (vf + v0) / 2 80m = (6m/s + v0)/2 First you would multiply both sides by 2 to eliminate the fraction, which then gives you 160m = 6m/s + v0. Now you would subtract 6 from both sides to give you 154m/s = v0 confidence assessment: 3
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22:24:43 `q004. We can reconcile the above solution with straightforward reasoning. How could the initial velocity have been reasoned out from the given information without the use of an equation? Hint: two of the quantities given in the problem can be combined to give another important quantity, which can then be combined with the third given quantity to reason out the final velocity.
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RESPONSE --> Without the equation you would obtain the answer in which I got which was 154 and I knew it didn't make any sense. Now I know. confidence assessment: 3
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22:35:59 `q005. Using the equation `ds = v0 `dt + .5 a `dt^2 determine the initial velocity of an object which accelerates uniformly at -2 m/s^2, starting at some unknown velocity, and is displaced 80 meters in 10 seconds. Begin by solving the equation for the unknown quantity and show every step.
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RESPONSE --> 'ds = v0('dt) + .5(a)('dt^2) 80m = v0(10s) + .5(-2m/s^2)(10s^2) 80m/v0 = 10s + -1m/s^2(100^2) 80m/v0 = 10s - 100m v0 = (10s-100m)/80m v0 = .125m/s - 1.25m^2 = -1.125 m/s This does not make any sense. confidence assessment: 1
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22:38:29 The change in the velocity of the object will be -2 m/s^2 * 10 s = -20 m/s. The object will therefore have a final velocity of 18 m/s - 20 m/s = -2 m/s. Its average velocity will be the average (18 m/s + (-2 m/s) ) / 2 = 8 m/s. An object which travels at an average velocity of 8 m/s for 10 sec will travel 80 meters.
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RESPONSE --> The change in the velocity of the object will be -2 m/s^2 * 10 s = -20 m/s. The object will therefore have a final velocity of 18 m/s - 20 m/s = -2 m/s. Its average velocity will be the average (18 m/s + (-2 m/s) ) / 2 = 8 m/s. An object which travels at an average velocity of 8 m/s for 10 sec will travel 80 meters. I understand what you are wanting now. self critique assessment: 3
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22:53:32 To solve for the unknown initial velocity v0 we start with vf^2 = v0^2 + 2 a `ds. We first add -2 a `ds to both sides to obtain vf^2 - 2 a `ds = v0^2. We then reverse the right-and left-hand sides and take the square root of both sides, obtaining v0 = +- `sqrt( vf^2 - 2 a `ds). We then substitute the given quantities vf = 20 m/s, `ds = 80 m and a = 3 m/s^2 to obtain v0 = +- `sqrt( (20 m/s)^2 - 2 * 2 m/s^2 * 80 m) = +- `sqrt( 400 m^2 / s^2 - 320 m^2 / s^2) = +- `sqrt(80 m^2 / s^2) = +- 8.9 m/s (approx.).
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RESPONSE --> vf^2 = v0^2 + 2(a)('ds) You would simply subtract v0^2 from both sides to get the equation 20m/s^2 - v0^2 = 2(2m/s^2)(80m); 400m^2/s^2 - v0^2 = 320m^2/s^2; v0 = -'sqrt(400m^2/s^2 - 320m^2/s^2); v0 = -'sqrt(80m^2/s^2) = 8.9m/s self critique assessment: 3
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23:08:58 In this case the average velocity will be ( -8.9 m/s + 20 m/s) / 2 = 5.5 m/s (approx) and the change in velocity will be 20 m/s - (-8.9 m/s) = 28.9 m/s (approx). At average velocity 5.5 meters/second the time required to displace the 80 meters will be 80 m / (5.5 sec) = 14.5 sec (approx). The velocity change of 28.5 meters/second in 14.5 sec implies an average acceleration of 28.5 m/s / (14.5 sec) = 2 m/s^2 (approx), again consistent with our results.
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RESPONSE --> When starting at -8.9ms we find the vAge would be (-8.9m/s + 20m/s)/2 = 5.5 m/s; The change in velocity would be 20m/s - -8.9m/s = 28.9m/s. And our time would be 80m / 5.5s = 14.5s. Now we can find the aAve by dividing 28.5m/s by 14.5s, which equals 2m/s^2 self critique assessment: 3
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23:11:43 `q009. Describe in commonsense terms the motion of the object in this example if its initial velocity is indeed -8.9 m/s. Assume that the object starts at the crossroads between two roads running North and South, and East and West, respectively, and that the object ends up 80 meters North of the crossroads. In what direction does it start out, what happens to its speed, and how does it end up where it does?
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RESPONSE --> The object would begin at the crossroads at a certain speed traveling North. Its speed is changing at a decreasing rate and it ends up at 80 meters North because it was at a certain speed and eventually slows at a rate of 8.9m/s and then stops at 80 meters north. confidence assessment: 3
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