course Phy 201 k݆ۈׂassignment #016
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11:23:09 `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 --> Since we know`ds,a,and v0 we can use the equation `ds=v0`dt+.5a`dt^2. `ds=2m,v0=0m/s and a=gravity 9.8m/s^2. Since v0=0 we can write the equation as `ds=.5a`dt^2, Rearrange the equation to find `dt^2=2`ds/a, take the sqrt from both sides, `dt=sqrt(2`ds/a),`dt=sqrt(2*2m/9.8m/s^2),`dt=sqrt(.41s),`dt=.63s confidence assessment: 3
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11:24:39 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 --> ok self critique assessment: 3
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11:27:00 `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 --> Sine we have `dt of .63s and a velocity of 12m/s then we can take v*`dt=`ds, `ds=12m/s*.63s,`ds=7.6m confidence assessment: 3
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11:27:13 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 --> ok self critique assessment: 3
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12:34:20 `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 --> Sinc it takes the same time going horizontal to the floor to reach as it is just dropped at the edge of the table we can fill in the equation `ds=v0`dt+2a`dt^2 to find the time. Since the intial v=0 we can eliminate the v0`dt., which gives us `ds=.5a`dt^2 we again arrange it to find `dt=sqrt(2`ds/a). We now`ds=1.5m,a=9.8m/s^2 so we have`dt=sqrt(2*1.5m/9.80m/s^2),`dt=sqrt(.31s),`dt=.55s. Since velocity does not change we can say `ds=6m/s*.55s,`ds=3.3m confidence assessment: 3
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12:35:58 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 --> ok self critique assessment: 3
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12:49:05 `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 --> As in the previous questions we can find out our time by using the vertical ball drop with v0=0. We again use the equation `dt=sqrt(2`ds/a). `dt=sqrt(2*4m/9.8m/s),`dt=sqrt(.82s),`dt=9s. Since we know `dt and horizontal distance of 32m we can now find avg horizontal velocity by taking `ds/`dt, 32m/.9s,vAve=35.6m/s and since the horizontal velocity does not change then vf=35.6m/s and v0=35.6m/s. To find the `dv we take a=9.8m/s^2 and `dt=.9s and multiply to get, 9.8m/s^2*.9s,`dv=8.8m/s. Since we know v0=0 then the vf=v0+`dv,vf=0m/s+8.8m/s,vf=8.8m/s confidence assessment: 2
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12:50:30 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 --> ok self critique assessment: 3
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