course phy 201 i am also including my answers to the practice quiz |K܇assignment #001 ֊DȘo Liberal Arts Mathematics I 02-20-2008
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16:30:54 Problem Number 1 A coasting cart has velocities 4.5 m/s, 10 m/s and 2.25 m/s at clock times t = 0 s, 9 s and 13 s. Approximately how far does it move and at what average rate does it accelerate over each of the two time intervals (i.e., 0- 9 seconds and 9 - 13 seconds)? Show how you could use a graph of velocity vs. time to obtain your results. Directly reason out your results using the concept of rate. If the automobile is coasting, and if air resistance is not a significant factor, then over which time interval do you think the average slope of the road is greatest?
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RESPONSE --> velocities=4.5m/s times 0s 10.m/s 9s 2.25m/s 13s 'ds=? aAve=? aAve='dv/'dt 'dv=vf-v0 10m/s-4.5m/s=5.5m/s 2.25m/s-10m/s=-7.75m/s 'dt=tf-t0 9s-0s=9s 13s-9s=4s 5.5m/s / 9s = .611m/s^2 = aAve (1) -7.75m/s 4s = -1.94m/s^2 = aAve (2) 'ds= vAve='ds/'dt vAve='dv/2 5.5m/s/2=2.75m/s = vAve (1) -7.75m/s/2=-3.88m/s = vAve (2) 2.75m/s = 'ds/9s 24.75m='ds -3.88m/s='ds/4s -15.52m='ds well the graph points would be (4.5,0) (10,9) (2.25,13) not sure about this direct reason thing. but lets see since the graph will show you the differece between the time intervals and the velocity intervals you can find the 'ds and 'dt points. and since vAve is the d's/'dt and aAve is the 'dv/'dt one can be derived by using the other. the slope is the rise/run and since the points are as follows 4.5/0 10/9 2.25/13 then i.d say the slope is greatest at point 2
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16:35:55 STUDENT ANSWER: The average acceleration for the cart from time 0s to 9s, the average velocity is (4.5+10)m/s/2=7.25 m/s so displacement is 7.25 m/s * 9s =65.25m. The displacement for the cart between clock time 9s and 13s is (2.25+10)m/s / 2 * 4s=24.5m The acceleration of the cart between clock time 0 - 9s is a = `dv / `dt ? (10-4.5)m/s / (9 s) =.6m/s^2. The acceleration of the cart between clock time 9-13s is (2.25-10( m/s / (4s) = -1.93m/^2. You could use a graph of velocity vs. time to obtain the results. By placing the information on a graph of v vs. t you could see the rise and the run between the points. For example looking at 0,4.5 and 9,10 you could see that it had a run of 9 s and a rise of 5.5 m/s and then to find the acceleration you would have ave accel = slope = 5.5 m/s / (9 s) = .61m/s^2. If the automobile is coasting then the slope of the road is the greatest when the magnitude of the acceleration is greatest. This occurs on the second slope, where the magnitude of the acceleration is 1.93 m/s^2. INSTRUCTOR NOTES FOR ALL STUDENTS: Be sure you use units correctly in every calculation, and be sure you are using the definitions of velocity (rate of change of position, so vAve = `ds / `dt) and acceleration (rate of change of velocity, so a = `dv / `dt).
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RESPONSE --> i think i have this
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16:59:21 Problem Number 2 We set up a straight ramp and measure the time required to coast down the ramp, from rest, at various slopes. The differences in elevation between one end and the other, for the different slopes, are 1.8, 4.2 and 6.7 cm. The time required for a cart to coast 78 cm down the ramp, starting from rest, is 2.722297 seconds on the first incline, 2.518101 seconds on the next, and 2.6606 seconds on the last. How well do these results support the hypothesis that acceleration is linearly dependent on slope?
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RESPONSE --> very well if we graph these points as 'ds vs 'dt then our points are (1.8,2.72) (4.2,2.51) (6.7,2.66) showing us that the larger increase in the elevation the faster our cart rolls down the slope. point 3 has a slope of 2.5.
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17:00:39 STUDENT RESPONSE: 1.8cm 2.722297s 4.2cm 2.518101s 6.7cm 2.6606s 'ds=78. These results show that the smallest slope the time to coast is the slowest The middle ramp has the fastest time down the ramp These results show that acceleration is fastest down the middle ramp (4.2cm) INSTRUCTOR COMMENT ON STUDENT RESPONSE: ** To support the idea that acceleration is linearly dependent on slope you first have to calculate the accelerations, as you did in the Video Experiments. You then need to plot a graph of acceleration vs. ramp slope and see if it forms a straight line. **
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RESPONSE --> ok so this one was wrong and defin. not thought out throughlly enough
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17:19:31 Problem Number 3 A ball reaches a ramp of length 50 cm with an unknown initial velocity and accelerates uniformly along the ramp, reaching the other end in 3.8 seconds. Its velocity at the end of the ramp is determined to be 6.31579 cm/s. What is its acceleration on the ramp?
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RESPONSE --> 'ds= 50cm vo= 0 'dt= 3.8s vf=6.32cm/s
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17:28:18 STUDENT RESPONSE: .'ds=50cm vf=6.31579cm/s 'dt = 3.8s The average velocity on the ramp is 50cm/3.8s =13.16m/s The initial velocity will be v0 = 2 'ds /dt, v0 = 2 *50 /3.8, v0 = 26.31579 -vf which is 6.31579, so the v0 is 20.00 ** You got it but this step isn?t correctly stated. v0 = 2 `ds/dt is not a correct statement. v0 = 2 vAve ? vf, or v0 = 2 `ds/`dt ? v0 (which is the same thing since vAve = `ds/`dt) is the correct equation. The reasoning is that ave velocity is the average of initial and final velocities, since acceleration is uniform. So you have vAve = (v0 + vf) / 2.. You know vAve from your previous calculation and vf is given. So you solve this equation for v0 and then substitute these values and simplify. You get v0 = 2 vAve ? vf, then substitute. This is actually what you did; just be careful you state it this way. If you don?t state it right there?s a chance you might not do it right. ** The accleration is 6.31579-20/3.8 = -3.601m/s/s If your v0 was correct this would be right **
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RESPONSE --> ok i didnt get it and now im really confused. if we have a uniformly accelerating item i thought we then were able to determine that v0=0.
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17:49:35 Problem Number 4 A projectile leaves the edge of a table at 18 cm/s, then falls freely a distance of 95 cm to the floor. How far does it travel in the horizontal direction during the fall?
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RESPONSE --> 'ds=95cm a=980cm/s^2 (gravity) v0=18cm/s the first way i opted was for the vf=sqrt equation but that didnt work gave me an vf of something like 430 cant be right right. but the details i have fit that equation so now i must find another set. i dont know the 'dt so that erraticas o lets see almost all the know equations. ok so lets say that vo is actually vf and we know a so lets solve for 'dt aAve='dv/'dt 980cm/s^2=18cm/s/'dt 980cm/s^2(dt)=18cm/s .02s='dt i have no idea where to proceed from here
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17:52:06 STUDENT RESPONSE: .'ds=95cm a=980cm/s/s v0=0 First, we take the equation 'ds = v0'dt + .5(980) 'dt^2 95cm = .5 (980) 'dt^2 'dt = sqrt 95cm/.5(980)=.194 So, 'dt = sqrt .194 'dt=.44s Next take the 18cm/s and multiply by the time duration .44s = 7.92cm/s INSTRUCTOR COMMENT TO ALL STUDENTS: Remember the basic principle used in this solution. You find the time of fall for the vertical motion, then find the horizontal displacement using the time interval and the initial horizontal velocity.
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RESPONSE --> ok so the 18cm/s was just filler information. until the end when it gets multiplied by teh 'dt that is found.
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18:33:22 Problem Number 5 What are the average acceleration and final velocity of an object which accelerates uniformly from rest, ending up with velocity 13.9 cm/sec after traveling a distance of 40 cm from start to finish?
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RESPONSE --> 'ds=40cm vf=13.9cm/s aAve=? vf=? first how can i find a velocity that is given to me in the equation? if we accelerate uniformly then we assume v0=0
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18:34:30 STUDENT RESPONSE: 'ds=40cm v0=0 vf=13.9cm/s If vf is 13.9cm/s then the using the equation a 'ds = vf^2 - v0^2)/2 will be , a * 40 cm = ( (13.9 cm/s)^2 ? 0^2) /2 40 * a = 193.21 cm^2/s^2 / 2 a = 96.61 cm^2/s^2 / (40 cm) a = 2.415cm/s^2 ** If vf^2 = v0^2 + 2 a `ds then a `ds = (vf^2 ? v0^2) / 2, as you say, so a = (vf^2 ? v0^2) / (2 `ds). This is what you did (good job) but be careful to state it this way. ** Then use the equation 'ds = v0 'dt + .5 a 'dt^2 ** I wouldn?t use this equation to solve for `dt, since the equation is quadratic in `dt. At this stage you know all variables except `dt. The equation vf = v0 + a `dt would be much simpler. Or you could just figure average velocity and divide into displacement. Either way would be fine. ** 40= +.5 2.415cm/s/s dt^2 dt = sqrt 40/ ** You could have used the fact that vAve = (13.9 cm/s + 0) / 2 = 6.95 cm/s to obtain `dt: Since vAve = `ds / `dt, we have `dt = `ds / vAve = 40 cm / (6.95 cm/s) = 5.8 sec, approx. **
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RESPONSE --> i started to use the 'ds=vAve'dt equation
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~RⵋþY| assignment #001 ֊DȘo Liberal Arts Mathematics I 02-20-2008 CԠCNvPyኦ}ಃ assignment # ֊DȘo Liberal Arts Mathematics I 02-20-2008 MNLꂖ~´~ assignment #008 008. `query 8 Physics I 02-20-2008
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18:39:21 QUESTION FROM STUDENT--Please define the differnece between Fnet and Force. See if you can answer this question.
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RESPONSE --> Force is any push or pull on any given object Fnet is the sum of all the forces upon said given object confidence assessment: 2
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18:39:45 ** Net force is the sum of all forces acting on an object. If you're pushing your car you are exerting a force, friction is opposing you, and the let force is the sum of the two (noting that one is positive, the other negative so you end up with net force less than the force you are exerting). Your heart rate responds to the force you are exerting and the speed with which the car is moving; the accel of the car depends on the net force. **
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RESPONSE --> ok self critique assessment: 2
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18:43:25 In terms of the equations of motion why do we expect that a * `ds is proportional to the change in v^2, and why do we then expect that the change in v^2 is proportional to Fnet `ds?
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RESPONSE --> im not sure confidence assessment: 0
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18:50:39 ** It's very important in physics to be able to think in terms of proportionality. To say that y is proportional to x is to say that for some k, y = k x. That is, y is a constant multiple of x. To say that a * `ds is proportional to the change in v^2 is to say that for some k, a * `ds = k * ( change in v^2)--i.e., that a * `ds is a constant multiple of the change in v^2. In terms of the equations of motion, we know that vf^2 = v0^2 + 2 a `ds so a `ds = 1/2 (vf^2 - v0^2), which is 1/2 the change in v^2. So a `ds is a constant multiple (1/2) of the change in v^2. Formally we have a `ds = k ( change in v^2) for k = 1/2. Now since Fnet = m a we see that Fnet is proportional to a for a given mass m, and it follows that Fnet `ds = k * change in v^2, for the appropriate k (specifically for k = mass / 2. **
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RESPONSE --> ok got it self critique assessment: 1
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18:51:54 How do our experimental results confirm or refute this hypothesis?
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RESPONSE --> our experiments when giving us our data to plug our equations either prove visually and tangibly what we suspect or they disprove them and suggest other explanations confidence assessment: 1
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18:52:24 ** We didn't actually do this part of the experiment, but on a ramp with fixed slope a `ds is simply proportional to `ds. When we measured `dt for different distances from rest down the same ramp, we were then able to determine the average and final velocities. The change in v^2 for each timing would be from 0 to vf^2. The change would therefore be just vf^2. If a `ds is proportional to the change in vf^2 then a graph of vf^2 vs. a `ds should be linear. Since a is constant we don't even need to determine it--a graph of vf^2 vs. `ds would be linear. This would confirm the hypothesis. **
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RESPONSE --> yes self critique assessment: 1
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Χⴿʸ assignment #008 008. `query 8 Physics I 02-20-2008"