course Phy 201 ???????????assignment #016
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18:20:53 `q001. Note that this assignment contains 4 questions. . How long does it take for an object dropped from rest to fall 2 meters under the influence of gravity?
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RESPONSE --> vo=0 a=9.8m/s^2 'ds=2m 'ds=vo*'dt+.5a'dt^2 Since vo=0, we areleft with 'ds=.5a'dt^2 Since we know 'ds but not 'dt, we can solve the equation for 'dt and fill in the variables that are known. 'dt=sqrt(2*2m/9.8m/s^2)=.64 seconds confidence assessment: 3
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18:21:03 The object has initial velocity 0, acceleration 9.8 meters/second^2, and displacement 2 meters. We can easily solve this problem using the equations of motion. Using the equation `ds = v0 `dt + .5 a `dt^2 we note first that since initial velocity is zero the term v0 `dt will be zero and can be dropped from the equation, leaving `ds = .5 a `dt^2. This equation is then easily solved for `dt, obtaining `dt = `sqrt(2 *`ds / a ) = `sqrt(2 * 2 meters / (9.8 m/s^2) ) = .64 second.
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RESPONSE --> I understand. self critique assessment: 3
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18:22:47 `q002. While an object dropped from rest falls 2 meters under the influence of gravity, another object moves along a level surface at 12 meters/second. How far does the second object move during the time required for the first object to fall?
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RESPONSE --> We already know 'dt from the previous problem, which was .64 seconds. We then just take the velocity and multiply it by the time to get the displacement. 12m/s*.64 seconds=7.7 meters confidence assessment: 3
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18:22:53 As we have seen in the preceding problem, the first object requires .64 second to fall. The second object will during this time move a distance of 12 meters/second * .64 second = 8 meters, approximately.
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RESPONSE --> I understand. self critique assessment: 3
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18:32:17 `q003. An object rolls off the edge of a tabletop and falls to the floor. At the instant it leaves the edge of the table is moving at 6 meters/second, and the distance from the tabletop to the floor is 1.5 meters. Since if we neglect air resistance there is zero net force in the horizontal direction, the horizontal velocity of the object will remain unchanged. Since the gravitational force acts in the vertical direction, motion in the vertical direction will undergo the acceleration of gravity. Since at the instant the object leaves the tabletop its motion is entirely in the horizontal direction, the vertical motion will also be characterized by an initial velocity of zero. How far will the object therefore travel in the horizontal direction before it strikes the floor?
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RESPONSE --> a=9.8 m/s^2 vertical 'ds=1.5m vo=0 'ds=.5a'dt^2 (vo is eliminated b/c it is 0). This is then reorganized and sovled for 'dt. 'dt=sqrt(2'ds/a)=sqrt(2*1.5m/9.8m/s^2)=.55 seconds When the object falls off the table and then hits the floor, the horizontal velocity from above will continue (6m/s). We can use the velocity formula to find the displacement since we know the time and the average velocity. 'ds=vAve/'dt =6m/s*.54 seconds =3.24 meters confidence assessment: 2
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18:32:25 We analyze the vertical motion first. The vertical motion is characterized by initial velocity zero, acceleration 9.8 meters/second^2 and displacement 1.5 meters. Since the initial vertical velocity is zero the equation `ds = v0 `dt + .5 a `dt^2 becomes `ds = .5 a `dt^2, which is easily solved for `dt to obtain `dt = `sqrt( 2 `ds / a) = `sqrt( 2 * 1.5 m / (9.8 m/s^2) ) = .54 sec, approx., so the object falls for about .54 seconds. The horizontal motion will therefore last .54 seconds. Since the initial 6 meter/second velocity is in the horizontal direction, and since the horizontal velocity is unchanging, the object will travel `ds = 6 m/s * .54 s = 3.2 m, approximately.
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RESPONSE --> Good, I got it. I understand this. self critique assessment: 3
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18:40:03 `q004. An object whose initial velocity is in the horizontal direction descends through a distance of 4 meters before it strikes the ground. It travels 32 meters in the horizontal direction during this time. What was its initial horizontal velocity? What are its final horizontal and vertical velocities?
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RESPONSE --> 'ds=4m a=9.8m/s^2 vo=0 'ds=vo*'dt+.5a'dt (vo is eliminated) solve for 'dt 'dt=sqrt (2'ds/a)=sqrt (2(4)/9.8m/s^2) =.82 seconds For the horizontal direction, the 'ds is 32 m and the time is still .82 seconds. So, the vAve horizontal velocity='ds/'dt =32m/.82 seconds =39 m/s The vertical velocity seems a bit more complicated. We know the aAve and 'dt, so we use this to find the average vertical velocity by multiplying the acceleration by the time. 9.8m/s^2*.82 seconds=8.04 meters/second. From this point, I am not sure how to get the initial and the final velocities. confidence assessment: 2
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18:41:54 We analyze the vertical motion first. The vertical motion is characterized by initial velocity zero, acceleration 9.8 meters/second^2 and displacement 4 meters. Since the initial vertical velocity is zero the equation `ds = v0 `dt + .5 a `dt^2 becomes `ds = .5 a `dt^2, which is easily solved for `dt to obtain `dt = `sqrt( 2 `ds / a) = `sqrt( 2 * 4 m / (9.8 m/s^2) ) = .9 sec, approx., so the object falls for about .9 seconds. The horizontal displacement during this .9 second fall is 32 meters, so the average horizontal velocity is 32 meters/(.9 second) = 35 meters/second, approximately. The final vertical velocity is easily calculated. The vertical velocity changes at a rate of 9.8 meters/second^2 for approximately .9 seconds, so the change in vertical velocity is `dv = 9.8 m/s^2 * .9 sec = 8.8 m/s. Since the initial vertical velocity was zero, final vertical velocity must be 8.8 meters/second in the downward direction. The final horizontal velocity is 35 meters/second, since the horizontal velocity remains unchanging until impact.
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RESPONSE --> Okay, I see that I just didn't go the last stepI I knew that vo=0, so I could have assumed that acceleration wouldbe uniform and would not change until the object changed from vertical to horizontal motion, meaning that the final velocity would actually be the velocity of the horizontal drop. self critique assessment: 2
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