course PHY201 匕ʰgλassignment #005 005. `query 5 Physics I 02-01-2008
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15:40:46 Intro Prob 6 Intro Prob 6 How do you find final velocity and displacement given initial velocity, acceleration and time interval?
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RESPONSE --> By substituting into equations: aAve = 'dv/'dt aAve*'dt = vf=v0 vf = aAve*'dt+vo confidence assessment: 3
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15:41:16 ** To find final velocity from the given quantities initial velocity, acceleration and `dt: Multiply `dt by accel to get `dv. Then add change in velocity `dv to init vel , and you have the final velocity**
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RESPONSE --> ok self critique assessment: 3
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15:41:44 Describe the flow diagram we obtain for the situation in which we know v0, vf and `dt.
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RESPONSE --> aAve = (vf-v0)/'dt confidence assessment: 3
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15:41:57 ** The flow diagram shows us the flow of information, what we get from what, usually by combining two quantites at a time. How we get each quantity may also be included. From vf and v0 we get `dv, shown by lines from vf and v0 at the top level to `dv. From vf and v0 we also get and vAve, shown by similar lines running from v0 and vf to vAve. Then from vAve and `dt we get `ds, with the accompanying lines indicating from vAve and `dt to `ds, while from `dv and `dt we get acceleration, indicated similarly. **
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RESPONSE --> ok self critique assessment: 3
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15:44:20 Principles of Physics and General College Physics Students: Prob. 1.26: Estimate how long it would take a runner at 10 km / hr to run from New York to California. Explain your solution thoroughly.
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RESPONSE --> Estimating that the journey would be 5000km/10km/hr = 500 hr confidence assessment: 3
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15:44:31 It is about 3000 miles from coast to coast. A km is about .62 mile, so 3000 miles * 1 km / (.62 miles) = 5000 km, approximately. At 10 km / hr, the time required would be 5000 km / (10 km / hr) = 500 km / (km/hr) = 500 km * (hr / km) = 500 (km / km) * hr = 500 hr.
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RESPONSE --> ok self critique assessment: 3
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15:46:04 All Students: Estimate the number heartbeats in a lifetime. What assumptions did you make to estimate the number of heartbeats in a human lifetime, and how did you obtain your final result?
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RESPONSE --> 50bpm*60min/hr*24hr/day*365day/yr*76yrs/lifetime = 2 million beats/lifetime confidence assessment: 3
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15:47:19 ** Typical assumptions: At 70 heartbeats per minute, with a lifetime of 80 years, we have 70 beats / minute * 60 minutes/hour * 24 hours / day * 365 days / year * 80 years = 3 billion, approximately. **
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RESPONSE --> dropped 3 zeros transferring solution to text box self critique assessment: 3
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15:47:27 University Physics Students Only: Problem 1.52 (i.e., Chapter 1, Problem 52): Angle between -2i+6j and 2i - 3j. What angle did you obtain between the two vectors?
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RESPONSE --> ok confidence assessment: 3
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15:47:32 ** For the given vectors we have dot product =-2 * 2 + 6 * (-3) = -22 magnitude of first vector = sqrt( (-2)^2 + 6^2) = sqrt(40) magnitude of second vector = sqrt( 2^2 + (-3)^2 ) = sqrt(13) Since dot product = magnitude of 1 st vector * magnitude of 2d vector * cos(theta) we have cos(theta) = dot product / (magnitude of 1 st vector * magnitude of 2d vector) so that theta = arccos [ dot product / (magnitude of 1 st vector * magnitude of 2d vector) ] = arccos[ -22 / ( sqrt(40) * sqrt(13) ) ] = arccos ( -.965) = 164 degrees, approx.. **
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RESPONSE --> ok self critique assessment: 3
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15:47:40 Add comments on any surprises or insights you experienced as a result of this assignment.
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RESPONSE --> none self critique assessment: 3
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15:47:47 ** I had to get a little help from a friend on vectors, but now I think I understand them. They are not as difficult to deal with as I thought. **
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RESPONSE --> ok self critique assessment: 3
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course PHY201 The program cut off before I could answer the last question -The object starts out at the intersection and originally travels south with a northward acceleration. Kind of like a car sliding backward on ice. Eventually the northward acceleration overtakes the southward movement and the object stops then starts moving northward.
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08:37:14 `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 --> 'dv=vf-v0=25-5=20m/s aAve='dv/'dt=20/4= 5m/s/s vAve=(vf+v0)/2=(25-5)/2 = 10m/s confidence assessment: 3
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08:37:43 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 --> Math error on my part self critique assessment: 3
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08:38:58 `q002. How far does the object of the preceding problem travel in the 4 seconds?
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RESPONSE --> vAve=(vf+v0)/2=(25+5)/2=15m/s 'ds=vAve*'dt = 15*4 = 60m confidence assessment: 3
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08:39:04 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 --> ok self critique assessment: 3
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08:50:38 `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 --> aAve = (vf-v0)/'dt vAve = (vf-v0)/2 'ds = vAve*'dt confidence assessment: 3
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09:08:48 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 self critique assessment: 3
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09:13:28 `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 --> see previous step confidence assessment: 3
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09:13:38 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 --> ok self critique assessment: 3
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09:13:50 `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 --> ok confidence assessment: 3
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09:14:01 When this average velocity is multiplied by `dt we get the displacement, which is `ds = (v0 + vf) / 2 * `dt.
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RESPONSE --> ok self critique assessment: 3
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09:16:28 `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 --> t,v 0,v0 - initial velovity 'dt,vf confidence assessment: 3
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09:17:44 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 thought you were asking for coordinate descriptions instead of the actual coordinates from the original problem self critique assessment: 3
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09:18:01 `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 --> increasing at a constant rate confidence assessment: 3
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09:18:09 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 self critique assessment: 3
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09:19:25 `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 --> slope is (vf-v0)/'dt = 5 = aAve confidence assessment: 3
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09:19:35 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 self critique assessment: 3
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09:20:59 `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 --> average altitude = (5+25)/2 = 15 = vAve A = 15*4 = 60 = 'ds confidence assessment: 3
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09:21:15 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 --> ok self critique assessment: 3
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Γ}T~sPօU assignment #006 006. Using equations with uniformly accelerated motion. Physics I 02-01-2008
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21:17:07 `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 --> aAve = (vf-v0)/'dt aAve = (30-10)/15 = 4/3 m/s/s confidence assessment: 3
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21:17:37 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 self critique assessment: 3
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21:18:32 `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 --> It is already known that a='dv/'dt confidence assessment: 3
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21:19:05 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 self critique assessment: 3
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21:25:39 `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 = 'dt(vf-v0)/2 2'ds = 'dt(vf-v0) 2'ds/'dt = vf-v0 v0 = vf - 2'ds/'dt v0 = 6-2*80/10 = -10m/s confidence assessment: 3
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21:26:23 We begin by solving the equation for v0. Starting with `ds = (vf + v0) / 2 * `dt, we can first multiply both sides of the equation by 2 / `dt, which gives us `ds * 2 / `dt = (vf + v0) / 2 * `dt * 2 / `dt. The right-hand side can be rearranged to give (vf + v0) * `dt / `dt * 2 / 2; since `dt / `dt = 1 and 2 / 2 = 1 the right-hand side becomes just vf + v0. The equation therefore becomes 2 * `ds / `dt = vf + v0. Adding -vf to both sides we obtain v0 = 2 * `ds / `dt - vf. We now plug in `ds = 80 meters, `dt = 10 sec and vf = 6 m/s to get v0 = 2 * 80 meters / 10 sec - 6 m/s = 160 meters / 10 sec - 6 m/s = 16 m/s - 6 m/s = 10 m/s.
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RESPONSE --> oops - dropped the negative self critique assessment: 3
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21:29:33 `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 --> vAve = 'ds/'dt = 80/10 = 8 (vf+vo)/2 = 8 (6+v0)/2 = 8 v0 = 10m/s confidence assessment: 3
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21:29:49 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 self critique assessment: 3
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21:34:26 `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+1/2(aAve)'dt^2 v0 = ('ds-0.5aAve'dt^2)/'dt v0 = (80-0.5(-2)10^2)/10 = 18 m/s confidence assessment: 3
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21:34:36 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 --> ok self critique assessment: 3
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21:38:04 `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 --> AN average velocity for this object would be 8m/s as the acceleration * time of acceleration yields -20m/s meaning the vo is -2m/s. Average velocity times time of motion = distance traveled - 80m confidence assessment: 3
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21:38:09 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 self critique assessment: 3
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21:41:35 `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 --> v0 = sqrt(vf^2-2aAve'ds) v0 = sqrt(20^2 - 2*2*80) = sqrt(80) m/s confidence assessment: 3
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21:41:43 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 --> ok self critique assessment: 3
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21:50:42 `q008. We can verify that starting at +8.9 m/s an object which attains a final velocity of 20 m/s while displacing 80 meters must accelerate at 2 m/s^2. In this case the average velocity will be ( 8.9 m/s + 20 m/s) / 2 = 14.5 m/s (approx) and the change in velocity will be 20 m/s - 8.9 m/s = 11.1 m/s. At average velocity 14.5 meters/second the time required to displace the 80 meters will be 80 m / (14.5 sec) = 5.5 sec (approx). The velocity change of 11.1 meters/second in 5.5 sec implies an average acceleration of 11.1 m/s / (5.5 sec) = 2 m/s^2 (approx), consistent with our results. Verify that starting at -8.9 m/s the object will also have acceleration 2 meters/second^2 if it ends up at velocity 20 m/s while displacing 80 meters.
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RESPONSE --> aAve = 'dv/'dt 'dt = 'dv/aAve = (vf-v0)/aAve 'dt = (20-(-8.9))/2 = 14.45s vAve = 'ds/'dt 'ds = vAve'dt = 'dt((vf+vo)/2) 'ds = 14.45((20+(-8.9))/2) = 80.198m confidence assessment: 3
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21:50:59 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 --> ok self critique assessment: 3
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