course phy 201 this did make more sense as i went.... ڮxǯt|Student Name:
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22:45:00 query Note that there are 10 questions in this assignment. At a certain instant the speedometer of a car reads 5 meters / second (of course cars in this country generally read speeds in miles per hour and km per hour, not meters / sec; but they could easily have their faces re-painted to read in meters/second, and we assume that this speedometer has been similarly altered). Exactly 4 seconds later the speedometer reads 25 meters/second (that, incidentally, indicates very good acceleration, as you will understand later). At what average rate is the speed of the car changing with respect to clock time?
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RESPONSE --> the car's speed changes from 5m/s to 25m/s over a 4 second period. In other words there is a difference of 20m/s over 4 seconds. 20m/s/4s= 5m/s/s
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22:45:22 The rate of change of the speed with respect clock time is equal to the change in the speed divided by the change in the clock time. So we must ask, what is the change in the speed, what is the change in the clock time and what therefore is the rate at which the speed is changing with respect to clock time? The change in speed from 5 meters/second to 25 meters/second is 20 meters/second. This occurs in a time interval lasting 4 seconds. The average rate of change of the speed is therefore (20 meters/second)/(4 seconds) = 5 meters / second / second. This means that on the average, per second, the speed changes by 5 meters/second.
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RESPONSE --> yep
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23:12:13 `q002. Explain in commonsense terms of the significance for an automobile of the rate at which its velocity changes. Do you think that a car with a more powerful engine would be capable of a greater rate of velocity change?
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RESPONSE --> Yes, I do believe that with cars comparable in weight and size a more powerful engine would be capable of a greater rate of velocity change. This rate is significant because it represents the acceleration of the vehicle which can determine whether or not you get smooshed by the semi behind you!!!
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23:12:47 A car whose velocity changes more rapidly will attain a given speed in a shorter time, and will be able to 'pull away from' a car which is capable of only a lesser rate of change in velocity. A more powerful engine, all other factors (e.g., weight and gearing) being equal, would be capable of a greater change in velocity in a given time interval.
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RESPONSE --> like I said....
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23:23:26 `q003. Explain how we obtain the units meters / second / second in our calculation of the rate of change of the car's speed.
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RESPONSE --> basically, you start by finding the difference in speed at the beginning and end of your measured time. So, for instance if you have a car that starts at 0 meters /second and ends up going 25 meters/second, then it has changed by 25 meters/second. Then, you divide this change over the time it took to acheive the change. If it took 5 seconds, then the speed changed by 25 meters/sec over 5 seconds. therefore it averaged a change of 5meters/sec every second of those 5 seconds. I say averaged, because any given second during that time period could be more or less, but the rate averages out to a change of 5 meters per second per second. the units are obtained by you form of measurement. obviously if you measured the speed of the car at 5 minute intervals, you would note the change in meters (or kilometres, or miles or whatever unit of distance you are measuring) per minute or you would have to convert the time unit back to seconds
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23:24:20 When we divide the change in velocity, expressed in meters/second, by the duration of the time interval in seconds, we get units of (meters / second) / second, often written meters / second / second.
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RESPONSE --> i think i wrote something similar--- yours is a lot more concise.......
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23:25:01 `q004. The unit (meters / second) / second is actually a complex fraction, having a numerator which is itself a fraction. Such a fraction can be simplified by multiplying the numerator by the reciprocal of the denominator. We thus get (meters / second) * (1/ second). What do we get when we multiply these two fractions?
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RESPONSE --> metres/sec^2
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23:25:22 Multiplying the numerators we get meters * 1; multiplying the denominators we get second * second, which is second^2. Our result is therefore meters * 1 / second^2, or just meters / second^2. If appropriate you may at this point comment on your understanding of the units of the rate of change of velocity.
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RESPONSE --> makes sense.
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23:27:14 `q004. If the velocity of an object changes from 10 m/s to -5 m/s during a time interval of 5 seconds, then at what average rate is the velocity changing?
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RESPONSE --> the object is slowing down 15m/s over a five second time period, so -15m/s/5s= -3m/s^2 or, it is slowing at the rate of 3m/s every second
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23:27:30 We see that the velocity changes from 10 meters/second to -5 meters/second, a change of -15 meters / second, during a five-second time interval. A change of -15 m/s during a 5 s time interval implies an average rate of -15 m/s / (5 s) = -3 (m/s)/ s = -3 m/s^2. This is the same as (-3 m/s) / s, as we saw above. So the velocity is changing by -3 m/s every second.
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RESPONSE --> yes
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23:28:24 `q005. You should have noted that velocity can be positive or negative, as can the change in velocity or the rate at which velocity changes. The average rate at which a quantity changes with respect to time over a given time interval is equal to the change in the quantity divided by the duration of the time interval. In this case we are calculating the average rate at which the velocity changes. If v represents velocity then we we use `dv to represent the change in velocity and `dt to represent the duration of the time interval. What expression do we therefore use to express the average rate at which the velocity changes?
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RESPONSE --> 'dv (change in velocity)/'dt (change in time)
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23:30:24 The average rate would be expressed by [ave rate of velocity change with respect to clock time] = `dv / `dt. The expression [ave rate of velocity change with respect to clock time] is pretty cumbersome so we give it a name. The name we give it is 'average acceleration', abbreviated by the letter aAve. Using a to represent acceleration, write down the definition of average acceleration. The definition of average acceleration is aAve = `dv / `dt. Please make any comments you feel appropriate about your understanding of the process so far.
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RESPONSE --> acceleration is the rate of change of velocity over time. aAve='dv/'dt
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23:32:49 `q006. If a runner is moving at 6 meters / sec at clock time t = 1.5 sec after starting a race, and at 9 meters / sec at clock time t = 3.5 sec after starting, then what is the average acceleration of the runner between these two clock times?
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RESPONSE --> First we find the change in velocity........ 9m/s- 6m/s= 3m/s Then we find the change in time..... 3.5 sec - 1.5 sec = 3 seconds. So the runner accelerates 3 m/sec over 3 seconds, or 3m/s/3s= 1 m/sec^2
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23:34:26 `q006a. What is the change `dv in the velocity of the runner during the time interval, and what is the change `dt in clock time during this interval?
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RESPONSE --> the change in velocity is 3 m/s (9m/s - 6m/s) and the change in time is 3 seconds (3.5 seconds - 1.5 seconds)
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23:35:39 `q006b. What therefore is the average rate at which the velocity is changing during this time interval?
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RESPONSE --> The average rate at which the velocity is changing during this interval is 3m/s over 3 seconds or 1m/s^2
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23:36:36 We see that the runner's velocity changes from 6 meters/second to 9 meters/second, a change of `dv = 9 m/s - 6 m/s = 3 m/s, during a time interval their runs from t = 1.5 sec to t = 3.5 sec so that the duration of the interval is `dt = 3.5 s - 1.5 s = 2.0 s. The rate at which the velocity changes is therefore 3 m/s / (2.0 s) = 1.5 m/s^2. Please comment if you wish on your understanding of the problem at this point.
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RESPONSE --> my understanding is fine---- now if I could just subtract 1 from 3.......
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23:42:00 `q007. On a graph of velocity vs. clock time, we can represent the two events of this problem by two points on the graph. The first point will be (1.5 sec, 6 meters/second) and the second point will be (3.5 sec, 9 meters / sec). What is the run between these points and what does it represent? What is the rise between these points what does it represent? What is the slope between these points what does it represent?
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RESPONSE --> the run, or horizontal distance is 2 (you notice that I subtracted correctly this time.....) and it represents the change in clock time. The rise, or vertical change is 3. This represents the change in velocity. the slope is found by dividing the rise by the run, or 3/2, this means that for every 2 places moved along the x axis of the graph, you move up 3 places along the y axis. On this particular graph, It means that for every 2 seconds, the velocity will change by 3m/s. so the slope represents the rate at which the velocity changes, or the acceleration.
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23:42:19 The rise from the first point to the second is from 6 meters/second to 9 meters/second, or 3 m/s. This represents the change `dv in velocity. The run is from 1.5 seconds to 3.5 seconds, or 2 seconds, and represents the change `dt in clock time. The slope, being the rise divided by the run, is 3 m/s / (2 sec) = 1.5 m/s^2. This slope represents `dv / `dt, which is the average acceleration during the time interval. You may if you wish comment on your understanding to this point.
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RESPONSE --> ok
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23:44:12 `q008. In what sense does the slope of any graph of velocity vs. clock time represent the acceleration of the object? For example, why does a greater slope imply greater acceleration?
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RESPONSE --> on any graph of velocity vs. clock time, the slope between any two clock times represents the rate at which the velocity changes during that time. If the slope is 0, there is no change in velocity. the steeper the slope, the more the velocity changes in a given time.
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23:44:25 Since the rise between two points on a graph of velocity vs. clock time represents the change in `dv velocity, and since the run represents the change `dt clock time, the slope represents rise / run, or change in velocity /change in clock time, or `dv / `dt. This is the definition of average acceleration.
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RESPONSE --> ok
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23:46:54 `q009. This is the same situation as in the preceding problem: An automobile coasts down a hill with a constant slope. At first its velocity increases at a very nearly constant rate. After it attains a certain velocity, air resistance becomes significant and the rate at which velocity changes decreases, though the velocity continues to increase. Describe a graph of velocity vs. clock time for this automobile (e.g., neither increasing nor decreasing; increasing at an increasing rate, a constant rate, a decreasing rate; decreasing at an increasing, constant or decreasing rate; the description could be different for different parts of the graph).
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RESPONSE --> This graph would be positively linear to a specific velocity. That is, until the given velocity, it would increase at a constant rate. However, at the point where the given velocity is reached, It would start increasing at a decreasing rate.
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23:48:55 Your graph should have velocity as the vertical axis and clock time as the horizontal axis. The graph should be increasing since the velocity starts at zero and increases. At first the graph should be increasing at a constant rate, because the velocity is increasing at a constant rate. The graph should continue increasing by after a time it should begin increasing at a decreasing rate, since the rate at which the velocity changes begins decreasing due to air resistance. However the graph should never decrease, although as air resistance gets greater and greater the graph might come closer and closer to leveling off. Critique your solution by describing or insights you had or insights you had and by explaining how you now know how to avoid those errors.
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RESPONSE --> I could have noted that increase could near zero in the second part of the graph, and it would have perhaps given a clearer picture of the graph.
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23:52:05 `q010. An automobile coasts down a hill with a constant slope. At first its velocity increases at a very nearly constant rate. After it attains a certain velocity, air resistance becomes significant and the rate at which velocity changes decreases, though the velocity continues to increase. Describe a graph of acceleration vs. clock time for this automobile (e.g., neither increasing nor decreasing; increasing at an increasing rate, a constant rate, a decreasing rate; decreasing at an increasing, constant or decreasing rate; the description could be different for different parts of the graph).
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RESPONSE --> Ok, for a second i thought you had written the same question twice. Actually, this graph will be flat (neither increasing nor decreasing) until the clock time where the given velocity is reached. At that point, the graph will decrease at a somewhat increasing rate until acceleration nears zero
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23:52:47 Your graph should have acceleration as the vertical axis and clock time as the horizontal axis. At first the graph should be neither increasing nor decreasing, since it first the acceleration is constant. Then after a time the graph should begin decreasing, which indicates the decreasing rate at which velocity changes as air resistance begins having an effect. An accurate description of whether the graph decreases at a constant, increasing or decreasing rate is not required at this point, because the reasoning is somewhat complex and requires knowledge you are not expected to possess at this point. However it is noted that the graph will at first decrease more and more rapidly, and then less and less rapidly as it approaches the t axis. In answer to the following question posed at this point by a student: Can you clarify some more the differences in acceleration and velocity? ** Velocity is the rate at which position changes and the standard units are cm/sec or m/sec. Acceleration is the rate at which velocity changes and its standard units are cm/s^2 or m/s^2. Velocity is the slope of a position vs. clock time graph. Acceleration is the slope of a velocity vs. clock time graph. **
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RESPONSE --> ok
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滯tUɴ~ Student Name: assignment #005 005. Uniformly Accelerated Motion
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08:59:31 `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 of the object changes by 20 m/s the average acceleration of the object is 20m/s/4s or 5 m/s/s the average velocity is the average of the two velocities given (5 + 25)/2= 15 m/s
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08:59:51 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).
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RESPONSE --> yes
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09:00:52 `q002. How far does the object of the preceding problem travel in the 4 seconds?
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RESPONSE --> well, with an average velocity of 15 m/s for 4 seconds, then the object should travel 60 m (4* 15)
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09:01:05 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 --> yes
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09:09:36 `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 --> we find the acceleration by figuring out the difference of the two velocities and then dividing that difference over the time it took to change to the final velocity. So, if an object takes 10 seconds to go from 0m/s to 60m/s, then it is accelerating 60/10 or 6ms every second (6m/s^2). We can figure out the distance traveled by finding the average velocity over that time. so we add 0 + 60 and divide by 2 to find the average velocity (in this case thirty) and multiply that by the time (10) so the object travels 300 m in 10 seconds
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09:09:46 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.
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RESPONSE --> ok
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09:11:56 `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.
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RESPONSE --> aAve= (vf-v0)/'dt vAve= (v0+vf)/2 displacement = (v0 + vf)/2 * 'dt
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09:12:11 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 --> yes
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09:12:24 `q005. The average velocity is the average of the initial and final velocities, which is expressed as (vf + v0) / 2.
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RESPONSE --> got that
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09:12:32 When this average velocity is multiplied by `dt we get the displacement, which is `ds = (v0 + vf) / 2 * `dt.
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RESPONSE --> yes
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10:14:38 `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?
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RESPONSE --> A graph of velocity vs clock time in a constant acceleration situation is linear, unless acceleration changes, there is no true final velocity. a clock time may be CHOSEN to represent vf for a particular problem (such as average velocity over a given interval) but no final velocity can be reached where acceleration continues. at T= 0 on the x axis, v0 is on the y axis at the intial velocity. vf would be the y coordinate corresponding to the velocity at the time the measurement stopped.
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10:15:35 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 --> ok, I assumed you were talking about a general graph of v vs t.
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10:17:03 `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 --> If the initial point is at (0, 5) and the the final point is at (4, 25) then the gaph is increasing at a constant rate.
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10:17:10 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).
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RESPONSE --> ok
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10:24:28 `q008. What is the slope of the graph between the two given points, and what is the meaning of this slope?
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RESPONSE --> the slope is rise/run or change in velocity over change in time, 25-5= 20, and 4-0 is 4, so the rise (change in velocity) over the run (change in time) is 5m/s/s which is the average acceleration over this time period.
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10:24:34 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 --> ok
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10:39:54 `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?
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RESPONSE --> the average altitude of the trapezoid is 15m/s [1/2 (a1 + a2)]. the average altitude (representing the average velocity during this time period) * the base (4s) gives us an area of 60m. this is the displacement of this object over the given time.
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10:40:09 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 --> yes
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qeDŎQȚ Student Name: assignment #006 006. Using equations with uniformly accelerated motion.
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01:10:06 `q001. Note that there are 9 questions in this assignment. Using the equation vf = v0 + a * `dt determine the acceleration of an object whose velocity increases at a uniform rate from 10 m/s to 30 m/s in 15 seconds. Begin by solving the equation for the acceleration a, then 'plug in' your initial and final velocities. Describe your work step y step.
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RESPONSE --> 30 m/s = 10 m/s + a(15 seconds) 20 m/s = a(15 seconds) 20m/s/15s= a= 11/3 m/s^2
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01:10:24 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 --> ok
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01:30:56 `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 --> we know that the velocity changes by 20 m/s (10 m/s to 20 m/s) and we know that the velocity changes over 15 seconds. so the velocity increases by 1 1/3 m/s every second.
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01:31:08 Knowing that acceleration is the average rate at which velocity changes, we would first find the change in velocity from 10 meters/second to 30 meters/second, which is 20 meters/second. We would then divided change in velocity by the time interval to get 20 meters/second / (15 sec) = 1.33 m/s^2.
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RESPONSE --> ok
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01:35:28 `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 --> 80m = (6m/s + v0)/2* 10s 80/10= 8 m/s = (6m/s + v0)/2 multiply both sides by 2 16m/s = 6m/s + v0 subtract 6m/s from both sides 10 m/s = v0 the object is slowing down.
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02:15:57 `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 --> once again, i have hit a button too soon.....
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02:16:08 The average velocity of the object is the average rate at which its position changes, which is equal to the 80 meters change in position divided by the 10 s change in clock time, or 80 meters / 10 sec = 8 meters / sec. Since the 8 m/s average velocity is equal to the average of the unknown initial velocity and the 6 m/s final velocity, we ask what quantity when average with 6 m/s will give us 8 m/s. Knowing that the average must be halfway between the two numbers being averaged, we see that the initial velocity must be 10 m/s. That is, 8 m/s is halfway between 6 m/s and 10 m/s.
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RESPONSE --> ok
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02:23:56 `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 --> 80 m = v0*10s + .5 * -2m/s^2*10^2 80m= v0*10s + .5 * -2m/s/s*100s2 simplify 80 m =10 v0s + (-1m/s^2 * 100s^2) 80m = v0 * 10s + -100 m add 100 m to both sides 180 m = v0 * 10s divide both sides by 10 s 18m/s= v0
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02:24:22 The unknown quantity is the initial velocity v0. To solve for v0 we start with `ds = v0 `dt + .5 a `dt^2. We first add -.5 a `dt^2 to both sides to obtain `ds - .5 a `dt^2 = v0 `dt. We then divide both sides by `dt to obtain (`ds - .5 a `dt^2) / `dt = v0. Then we substitute the given displacement `ds = 80 meters, acceleration a = -2 m/s^2 and time interval `dt = 10 seconds to obtain v0 = [ 80 meters - .5 * (-2 m/s^2) * (10 sec)^2 ] / (10 sec) = [ 80 meters - .5 * (-2 m/s^2) * 100 s^2 ] / (10 sec) = [ 80 m - (-100 m) ] / (10 sec) = 180 m / (10 s) = 18 m/s.
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RESPONSE --> i went about it a little differently.....
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21:26:31 `q006. Check the consistency of this result by verifying, by direct reasoning rather than equations, that an object whose initial velocity is 18 m/s and which accelerates for 10 seconds at an acceleration of -2 m/s^2 does indeed experience a displacement of 80 meters.
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RESPONSE --> - 2m/s^2 for 10 seconds is a change in velocity of -20 m/s over 10 seconds. since the initial velocity is 18m/s, then - 20 m/s would give us a final velocity of -2 m/s at the end of 10 seconds. this gives us an average velocity of (18m/s + -2 m/s)/2, or 16m/s/2or 8 m/s. 8m/s * 10 seconds gives us 80 meters.
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21:26:46 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 --> ok
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21:37:40 `q007. Using the equation vf^2 = v0^2 + 2 a `ds determine the initial velocity of an object which attains a final velocity of 20 meters/second after accelerating uniformly at 2 meters/second^2 through a displacement of 80 meters. Begin by solving the equation for the unknown quantity and show every step.
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RESPONSE --> (20 m/s)^2= v0^2 + 2 * 2m/s^s * 80m. 400 m/s^2= v0^2 + 4 m/s^2 * 80m 400 m^2/s^2 = v0^2 + 320 m^2/s^2 400 m2/s2- 320 m2/s2= v02 + 320m2/s2 -320m2/s2 80m^2/s^2= v0^2 square root each side 8.944 m/s = v0
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21:38:41 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 --> i seem to take the long way around in these problems
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"
course phy 201 this did make more sense as i went.... ڮxǯt|Student Name:
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22:45:00 query Note that there are 10 questions in this assignment. At a certain instant the speedometer of a car reads 5 meters / second (of course cars in this country generally read speeds in miles per hour and km per hour, not meters / sec; but they could easily have their faces re-painted to read in meters/second, and we assume that this speedometer has been similarly altered). Exactly 4 seconds later the speedometer reads 25 meters/second (that, incidentally, indicates very good acceleration, as you will understand later). At what average rate is the speed of the car changing with respect to clock time?
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RESPONSE --> the car's speed changes from 5m/s to 25m/s over a 4 second period. In other words there is a difference of 20m/s over 4 seconds. 20m/s/4s= 5m/s/s
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22:45:22 The rate of change of the speed with respect clock time is equal to the change in the speed divided by the change in the clock time. So we must ask, what is the change in the speed, what is the change in the clock time and what therefore is the rate at which the speed is changing with respect to clock time? The change in speed from 5 meters/second to 25 meters/second is 20 meters/second. This occurs in a time interval lasting 4 seconds. The average rate of change of the speed is therefore (20 meters/second)/(4 seconds) = 5 meters / second / second. This means that on the average, per second, the speed changes by 5 meters/second.
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RESPONSE --> yep
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23:12:13 `q002. Explain in commonsense terms of the significance for an automobile of the rate at which its velocity changes. Do you think that a car with a more powerful engine would be capable of a greater rate of velocity change?
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RESPONSE --> Yes, I do believe that with cars comparable in weight and size a more powerful engine would be capable of a greater rate of velocity change. This rate is significant because it represents the acceleration of the vehicle which can determine whether or not you get smooshed by the semi behind you!!!
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23:12:47 A car whose velocity changes more rapidly will attain a given speed in a shorter time, and will be able to 'pull away from' a car which is capable of only a lesser rate of change in velocity. A more powerful engine, all other factors (e.g., weight and gearing) being equal, would be capable of a greater change in velocity in a given time interval.
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RESPONSE --> like I said....
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23:23:26 `q003. Explain how we obtain the units meters / second / second in our calculation of the rate of change of the car's speed.
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RESPONSE --> basically, you start by finding the difference in speed at the beginning and end of your measured time. So, for instance if you have a car that starts at 0 meters /second and ends up going 25 meters/second, then it has changed by 25 meters/second. Then, you divide this change over the time it took to acheive the change. If it took 5 seconds, then the speed changed by 25 meters/sec over 5 seconds. therefore it averaged a change of 5meters/sec every second of those 5 seconds. I say averaged, because any given second during that time period could be more or less, but the rate averages out to a change of 5 meters per second per second. the units are obtained by you form of measurement. obviously if you measured the speed of the car at 5 minute intervals, you would note the change in meters (or kilometres, or miles or whatever unit of distance you are measuring) per minute or you would have to convert the time unit back to seconds
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23:24:20 When we divide the change in velocity, expressed in meters/second, by the duration of the time interval in seconds, we get units of (meters / second) / second, often written meters / second / second.
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RESPONSE --> i think i wrote something similar--- yours is a lot more concise.......
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23:25:01 `q004. The unit (meters / second) / second is actually a complex fraction, having a numerator which is itself a fraction. Such a fraction can be simplified by multiplying the numerator by the reciprocal of the denominator. We thus get (meters / second) * (1/ second). What do we get when we multiply these two fractions?
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RESPONSE --> metres/sec^2
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23:25:22 Multiplying the numerators we get meters * 1; multiplying the denominators we get second * second, which is second^2. Our result is therefore meters * 1 / second^2, or just meters / second^2. If appropriate you may at this point comment on your understanding of the units of the rate of change of velocity.
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RESPONSE --> makes sense.
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23:27:14 `q004. If the velocity of an object changes from 10 m/s to -5 m/s during a time interval of 5 seconds, then at what average rate is the velocity changing?
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RESPONSE --> the object is slowing down 15m/s over a five second time period, so -15m/s/5s= -3m/s^2 or, it is slowing at the rate of 3m/s every second
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23:27:30 We see that the velocity changes from 10 meters/second to -5 meters/second, a change of -15 meters / second, during a five-second time interval. A change of -15 m/s during a 5 s time interval implies an average rate of -15 m/s / (5 s) = -3 (m/s)/ s = -3 m/s^2. This is the same as (-3 m/s) / s, as we saw above. So the velocity is changing by -3 m/s every second.
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RESPONSE --> yes
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23:28:24 `q005. You should have noted that velocity can be positive or negative, as can the change in velocity or the rate at which velocity changes. The average rate at which a quantity changes with respect to time over a given time interval is equal to the change in the quantity divided by the duration of the time interval. In this case we are calculating the average rate at which the velocity changes. If v represents velocity then we we use `dv to represent the change in velocity and `dt to represent the duration of the time interval. What expression do we therefore use to express the average rate at which the velocity changes?
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RESPONSE --> 'dv (change in velocity)/'dt (change in time)
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23:30:24 The average rate would be expressed by [ave rate of velocity change with respect to clock time] = `dv / `dt. The expression [ave rate of velocity change with respect to clock time] is pretty cumbersome so we give it a name. The name we give it is 'average acceleration', abbreviated by the letter aAve. Using a to represent acceleration, write down the definition of average acceleration. The definition of average acceleration is aAve = `dv / `dt. Please make any comments you feel appropriate about your understanding of the process so far.
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RESPONSE --> acceleration is the rate of change of velocity over time. aAve='dv/'dt
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23:32:49 `q006. If a runner is moving at 6 meters / sec at clock time t = 1.5 sec after starting a race, and at 9 meters / sec at clock time t = 3.5 sec after starting, then what is the average acceleration of the runner between these two clock times?
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RESPONSE --> First we find the change in velocity........ 9m/s- 6m/s= 3m/s Then we find the change in time..... 3.5 sec - 1.5 sec = 3 seconds. So the runner accelerates 3 m/sec over 3 seconds, or 3m/s/3s= 1 m/sec^2
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23:34:26 `q006a. What is the change `dv in the velocity of the runner during the time interval, and what is the change `dt in clock time during this interval?
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RESPONSE --> the change in velocity is 3 m/s (9m/s - 6m/s) and the change in time is 3 seconds (3.5 seconds - 1.5 seconds)
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23:35:39 `q006b. What therefore is the average rate at which the velocity is changing during this time interval?
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RESPONSE --> The average rate at which the velocity is changing during this interval is 3m/s over 3 seconds or 1m/s^2
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23:36:36 We see that the runner's velocity changes from 6 meters/second to 9 meters/second, a change of `dv = 9 m/s - 6 m/s = 3 m/s, during a time interval their runs from t = 1.5 sec to t = 3.5 sec so that the duration of the interval is `dt = 3.5 s - 1.5 s = 2.0 s. The rate at which the velocity changes is therefore 3 m/s / (2.0 s) = 1.5 m/s^2. Please comment if you wish on your understanding of the problem at this point.
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RESPONSE --> my understanding is fine---- now if I could just subtract 1 from 3.......
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23:42:00 `q007. On a graph of velocity vs. clock time, we can represent the two events of this problem by two points on the graph. The first point will be (1.5 sec, 6 meters/second) and the second point will be (3.5 sec, 9 meters / sec). What is the run between these points and what does it represent? What is the rise between these points what does it represent? What is the slope between these points what does it represent?
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RESPONSE --> the run, or horizontal distance is 2 (you notice that I subtracted correctly this time.....) and it represents the change in clock time. The rise, or vertical change is 3. This represents the change in velocity. the slope is found by dividing the rise by the run, or 3/2, this means that for every 2 places moved along the x axis of the graph, you move up 3 places along the y axis. On this particular graph, It means that for every 2 seconds, the velocity will change by 3m/s. so the slope represents the rate at which the velocity changes, or the acceleration.
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23:42:19 The rise from the first point to the second is from 6 meters/second to 9 meters/second, or 3 m/s. This represents the change `dv in velocity. The run is from 1.5 seconds to 3.5 seconds, or 2 seconds, and represents the change `dt in clock time. The slope, being the rise divided by the run, is 3 m/s / (2 sec) = 1.5 m/s^2. This slope represents `dv / `dt, which is the average acceleration during the time interval. You may if you wish comment on your understanding to this point.
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RESPONSE --> ok
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23:44:12 `q008. In what sense does the slope of any graph of velocity vs. clock time represent the acceleration of the object? For example, why does a greater slope imply greater acceleration?
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RESPONSE --> on any graph of velocity vs. clock time, the slope between any two clock times represents the rate at which the velocity changes during that time. If the slope is 0, there is no change in velocity. the steeper the slope, the more the velocity changes in a given time.
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23:44:25 Since the rise between two points on a graph of velocity vs. clock time represents the change in `dv velocity, and since the run represents the change `dt clock time, the slope represents rise / run, or change in velocity /change in clock time, or `dv / `dt. This is the definition of average acceleration.
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RESPONSE --> ok
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23:46:54 `q009. This is the same situation as in the preceding problem: An automobile coasts down a hill with a constant slope. At first its velocity increases at a very nearly constant rate. After it attains a certain velocity, air resistance becomes significant and the rate at which velocity changes decreases, though the velocity continues to increase. Describe a graph of velocity vs. clock time for this automobile (e.g., neither increasing nor decreasing; increasing at an increasing rate, a constant rate, a decreasing rate; decreasing at an increasing, constant or decreasing rate; the description could be different for different parts of the graph).
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RESPONSE --> This graph would be positively linear to a specific velocity. That is, until the given velocity, it would increase at a constant rate. However, at the point where the given velocity is reached, It would start increasing at a decreasing rate.
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23:48:55 Your graph should have velocity as the vertical axis and clock time as the horizontal axis. The graph should be increasing since the velocity starts at zero and increases. At first the graph should be increasing at a constant rate, because the velocity is increasing at a constant rate. The graph should continue increasing by after a time it should begin increasing at a decreasing rate, since the rate at which the velocity changes begins decreasing due to air resistance. However the graph should never decrease, although as air resistance gets greater and greater the graph might come closer and closer to leveling off. Critique your solution by describing or insights you had or insights you had and by explaining how you now know how to avoid those errors.
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RESPONSE --> I could have noted that increase could near zero in the second part of the graph, and it would have perhaps given a clearer picture of the graph.
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23:52:05 `q010. An automobile coasts down a hill with a constant slope. At first its velocity increases at a very nearly constant rate. After it attains a certain velocity, air resistance becomes significant and the rate at which velocity changes decreases, though the velocity continues to increase. Describe a graph of acceleration vs. clock time for this automobile (e.g., neither increasing nor decreasing; increasing at an increasing rate, a constant rate, a decreasing rate; decreasing at an increasing, constant or decreasing rate; the description could be different for different parts of the graph).
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RESPONSE --> Ok, for a second i thought you had written the same question twice. Actually, this graph will be flat (neither increasing nor decreasing) until the clock time where the given velocity is reached. At that point, the graph will decrease at a somewhat increasing rate until acceleration nears zero
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23:52:47 Your graph should have acceleration as the vertical axis and clock time as the horizontal axis. At first the graph should be neither increasing nor decreasing, since it first the acceleration is constant. Then after a time the graph should begin decreasing, which indicates the decreasing rate at which velocity changes as air resistance begins having an effect. An accurate description of whether the graph decreases at a constant, increasing or decreasing rate is not required at this point, because the reasoning is somewhat complex and requires knowledge you are not expected to possess at this point. However it is noted that the graph will at first decrease more and more rapidly, and then less and less rapidly as it approaches the t axis. In answer to the following question posed at this point by a student: Can you clarify some more the differences in acceleration and velocity? ** Velocity is the rate at which position changes and the standard units are cm/sec or m/sec. Acceleration is the rate at which velocity changes and its standard units are cm/s^2 or m/s^2. Velocity is the slope of a position vs. clock time graph. Acceleration is the slope of a velocity vs. clock time graph. **
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RESPONSE --> ok
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滯tUɴ~ Student Name: assignment #005 005. Uniformly Accelerated Motion
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08:59:31 `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 of the object changes by 20 m/s the average acceleration of the object is 20m/s/4s or 5 m/s/s the average velocity is the average of the two velocities given (5 + 25)/2= 15 m/s
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08:59:51 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).
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RESPONSE --> yes
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09:00:52 `q002. How far does the object of the preceding problem travel in the 4 seconds?
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RESPONSE --> well, with an average velocity of 15 m/s for 4 seconds, then the object should travel 60 m (4* 15)
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09:01:05 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 --> yes
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09:09:36 `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 --> we find the acceleration by figuring out the difference of the two velocities and then dividing that difference over the time it took to change to the final velocity. So, if an object takes 10 seconds to go from 0m/s to 60m/s, then it is accelerating 60/10 or 6ms every second (6m/s^2). We can figure out the distance traveled by finding the average velocity over that time. so we add 0 + 60 and divide by 2 to find the average velocity (in this case thirty) and multiply that by the time (10) so the object travels 300 m in 10 seconds
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09:09:46 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.
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RESPONSE --> ok
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09:11:56 `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.
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RESPONSE --> aAve= (vf-v0)/'dt vAve= (v0+vf)/2 displacement = (v0 + vf)/2 * 'dt
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09:12:11 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 --> yes
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09:12:24 `q005. The average velocity is the average of the initial and final velocities, which is expressed as (vf + v0) / 2.
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RESPONSE --> got that
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09:12:32 When this average velocity is multiplied by `dt we get the displacement, which is `ds = (v0 + vf) / 2 * `dt.
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RESPONSE --> yes
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10:14:38 `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?
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RESPONSE --> A graph of velocity vs clock time in a constant acceleration situation is linear, unless acceleration changes, there is no true final velocity. a clock time may be CHOSEN to represent vf for a particular problem (such as average velocity over a given interval) but no final velocity can be reached where acceleration continues. at T= 0 on the x axis, v0 is on the y axis at the intial velocity. vf would be the y coordinate corresponding to the velocity at the time the measurement stopped.
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10:15:35 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 --> ok, I assumed you were talking about a general graph of v vs t.
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10:17:03 `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 --> If the initial point is at (0, 5) and the the final point is at (4, 25) then the gaph is increasing at a constant rate.
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10:17:10 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).
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RESPONSE --> ok
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10:24:28 `q008. What is the slope of the graph between the two given points, and what is the meaning of this slope?
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RESPONSE --> the slope is rise/run or change in velocity over change in time, 25-5= 20, and 4-0 is 4, so the rise (change in velocity) over the run (change in time) is 5m/s/s which is the average acceleration over this time period.
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10:24:34 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 --> ok
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10:39:54 `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?
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RESPONSE --> the average altitude of the trapezoid is 15m/s [1/2 (a1 + a2)]. the average altitude (representing the average velocity during this time period) * the base (4s) gives us an area of 60m. this is the displacement of this object over the given time.
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10:40:09 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 --> yes
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qeDŎQȚ Student Name: assignment #006 006. Using equations with uniformly accelerated motion.
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01:10:06 `q001. Note that there are 9 questions in this assignment. Using the equation vf = v0 + a * `dt determine the acceleration of an object whose velocity increases at a uniform rate from 10 m/s to 30 m/s in 15 seconds. Begin by solving the equation for the acceleration a, then 'plug in' your initial and final velocities. Describe your work step y step.
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RESPONSE --> 30 m/s = 10 m/s + a(15 seconds) 20 m/s = a(15 seconds) 20m/s/15s= a= 11/3 m/s^2
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01:10:24 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 --> ok
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01:30:56 `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 --> we know that the velocity changes by 20 m/s (10 m/s to 20 m/s) and we know that the velocity changes over 15 seconds. so the velocity increases by 1 1/3 m/s every second.
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01:31:08 Knowing that acceleration is the average rate at which velocity changes, we would first find the change in velocity from 10 meters/second to 30 meters/second, which is 20 meters/second. We would then divided change in velocity by the time interval to get 20 meters/second / (15 sec) = 1.33 m/s^2.
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RESPONSE --> ok
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01:35:28 `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 --> 80m = (6m/s + v0)/2* 10s 80/10= 8 m/s = (6m/s + v0)/2 multiply both sides by 2 16m/s = 6m/s + v0 subtract 6m/s from both sides 10 m/s = v0 the object is slowing down.
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02:15:57 `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 --> once again, i have hit a button too soon.....
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02:16:08 The average velocity of the object is the average rate at which its position changes, which is equal to the 80 meters change in position divided by the 10 s change in clock time, or 80 meters / 10 sec = 8 meters / sec. Since the 8 m/s average velocity is equal to the average of the unknown initial velocity and the 6 m/s final velocity, we ask what quantity when average with 6 m/s will give us 8 m/s. Knowing that the average must be halfway between the two numbers being averaged, we see that the initial velocity must be 10 m/s. That is, 8 m/s is halfway between 6 m/s and 10 m/s.
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RESPONSE --> ok
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02:23:56 `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 --> 80 m = v0*10s + .5 * -2m/s^2*10^2 80m= v0*10s + .5 * -2m/s/s*100s2 simplify 80 m =10 v0s + (-1m/s^2 * 100s^2) 80m = v0 * 10s + -100 m add 100 m to both sides 180 m = v0 * 10s divide both sides by 10 s 18m/s= v0
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02:24:22 The unknown quantity is the initial velocity v0. To solve for v0 we start with `ds = v0 `dt + .5 a `dt^2. We first add -.5 a `dt^2 to both sides to obtain `ds - .5 a `dt^2 = v0 `dt. We then divide both sides by `dt to obtain (`ds - .5 a `dt^2) / `dt = v0. Then we substitute the given displacement `ds = 80 meters, acceleration a = -2 m/s^2 and time interval `dt = 10 seconds to obtain v0 = [ 80 meters - .5 * (-2 m/s^2) * (10 sec)^2 ] / (10 sec) = [ 80 meters - .5 * (-2 m/s^2) * 100 s^2 ] / (10 sec) = [ 80 m - (-100 m) ] / (10 sec) = 180 m / (10 s) = 18 m/s.
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RESPONSE --> i went about it a little differently.....
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21:26:31 `q006. Check the consistency of this result by verifying, by direct reasoning rather than equations, that an object whose initial velocity is 18 m/s and which accelerates for 10 seconds at an acceleration of -2 m/s^2 does indeed experience a displacement of 80 meters.
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RESPONSE --> - 2m/s^2 for 10 seconds is a change in velocity of -20 m/s over 10 seconds. since the initial velocity is 18m/s, then - 20 m/s would give us a final velocity of -2 m/s at the end of 10 seconds. this gives us an average velocity of (18m/s + -2 m/s)/2, or 16m/s/2or 8 m/s. 8m/s * 10 seconds gives us 80 meters.
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21:26:46 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 --> ok
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21:37:40 `q007. Using the equation vf^2 = v0^2 + 2 a `ds determine the initial velocity of an object which attains a final velocity of 20 meters/second after accelerating uniformly at 2 meters/second^2 through a displacement of 80 meters. Begin by solving the equation for the unknown quantity and show every step.
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RESPONSE --> (20 m/s)^2= v0^2 + 2 * 2m/s^s * 80m. 400 m/s^2= v0^2 + 4 m/s^2 * 80m 400 m^2/s^2 = v0^2 + 320 m^2/s^2 400 m2/s2- 320 m2/s2= v02 + 320m2/s2 -320m2/s2 80m^2/s^2= v0^2 square root each side 8.944 m/s = v0
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21:38:41 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 --> i seem to take the long way around in these problems
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