course phy good luck! òæ×;َËÚ‰Ê|ízäñàCÎ Q¡îîâìassignment #004
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15:18:40 How do we reason out the process of determining acceleration from rest given displacement and time duration?
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RESPONSE --> If an obect starts from rest, and travels a given distance in a given amount of time, then we canfind the acceleration. first, we find the average velocity by dividing distance traveled by the time (ie, a ball rolls 15 cm in 3 sec--> 15/3= 5 cm/sec) then, knowing that the average velocity is the final velocity/2 when an object starts at rest [ (vf + v0)/2 = vave ---> (vf + 0)/2=5 cm/sec --> vf/2 = 5 cm/sec --> vf= 10cm/sec] we find the final velocity-- in this case 10 cm/sec. the acceleration is the difference velocity over time, so (vf-v0)/t in the example, (10 cm/sec - 0 cm/sec)/3 sec = 3.33 cm/sec/sec
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15:19:22 ** COMMON ERROR: By first finding the average velocity. Also we must understand this gives us the change in velocity which gives us the acceleration. INSTRUCTOR COMMENT: Average acceleration is not related to average velocity. It is related to the change in velocity, but change in velocity is not the same thing as acceleration. Acceleration is the rate of change of velocity. ANSWER TO QUESTION: Average acceleration is defined to be the average rate at which velocity changes, which is change in velocity/change in clock time, or equivalently change in velocity/time duration. However we do not know the change in velocity, nor from the given information can we directly determine the change in velocity. So we have to look at what we can reason out from what we know. Given displacement from rest and time duration we calculate average velocity: vAve = `ds / `dt = displacement/time duration. Since the initial velocity is 0, the final velocity is double the average so we find the final velocity by doubling the average velocity. Now that we know the initial velocity 0 and the final velocity we can find change in velocity by subtracting initial vel from final vel. Dividing change in velocity by a time duration we finally obtain the average acceleration. **
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RESPONSE --> exactly
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¯±ç×Ø¿}Çy䌀cÉxÅòÉ¿ËðˆPïóÇʲD± assignment # òæ×;َËÚ‰Ê|ízäñàCÎ Q¡îîâì Liberal Arts Mathematics I 08-03-2006
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21:35:29 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 --> ok
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23:24:54 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 --> ok, with ramp elevated 1.8 cm the average velocity is 78cm/2.722297 or 28.7 cm/sec associated with a slope created by a right sided triangle with height 1.8 and hypotenuse 78 cm-- 1.8^2 + width^2 = 78^2 which is 78^2-1.8^2 = w^2 = 77.97cm 1.8/77.97= a slope of .023 cm. 28.7 cm/sec slope of .023cm with ramp elevated 4.2 cm, the average velocity is 78cm/2.518101 sec or 30.97572337cm/sec ~ 31cm/sec the slope here is 78^2- 4.2^2= w^2 w=77.88 rise over run = 4.2/77.88 or .052 cm slope 30.9757cm/sec with 6.7 cm ramp, the average velocity 78cm/2.6606sec = 29.31669548 cm/sec ~ 29.3 cm/sec the slope is 78^2-6.7^2= 77.71^2 rise over run= 6.7/77.71 or a slope of .086 cm slope with a velocity of 29.3 cm/sec since the highest velocity is actually associated with a lower slope, the data does not support the hypopthesis
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23:25:43 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 -->
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23:26:18 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 --> it doesn't form a line
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23:58:58 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 --> since vave * time= distance, and since vave= (vf+v0)/2, then (6.31579 + v0)/2*(3.8)= 50cm 13.1579 cm/s= (6.31579 cm/s + v0)/2 26.3158 cm/s - 6.31579 = v0 19.9999~ 20 cm/sec acceleration is the (vf-v0)/3.8 seconds in this case there is negative acceleration -3.6011 cm/sec^2
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00:00:40 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 --> Enter, as appropriate, an answer to the question, a critique of your answer in response to a given answer, your insights regarding the situation at this point, notes to yourself, or just an OK. Always critique your solutions by describing any insights you had or errors you makde, and by explaining how you can make use of the insight or how you now know how to avoid certain errors. Also pose for the instructor any question or questions that you have related to the problem or series of problems. yes.
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07:00:24 How do we obtain an approximate velocity vs. clock time table and graph from a series of position vs. clock time observations? How do we then obtain an approximate acceleration vs. clock timetable and graph?
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RESPONSE --> If we have a graph of position vs. clock time, then the slope of the graph between any two clock times represents the average velocity between those times. by measuring the slope at short time intervals, we can then recreate (a fairly accurate) velocity vs time graph.
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07:01:25 ** If we start with a position vs. clock time graph we do divide the graph into intervals. On each interval velocity, which is rate of change of position, is represented by the slope of the graph. So we have to calculate the slope of the graph on each interval. If we then graph the slopes vs. the midpoint times of the intervals we get a good approximation of the velocity vs. clock time graph. The velocity at a given instant is the slope of the position vs. clock time graph. STUDENT ANSWER: We first calculate the time interval and displacement between each pair of data points. We use these calculations to calculate the average velocity. To obtain an approximate accelerations. Clock timetable and graph by associating the average velocity on a time interval with the midpoint clock time on that interval. INSTRUCTOR CRITIQUE: Using your words and amplifying a bit: We first calculate the time interval and displacement between each pair of data points. We use these results to calculate the average velocity, dividing displacement by time interval for each interval. Then we make a table, showing the average velocity vs. the midpoint time for each time interval. To obtain approximate accelerations we use the table and graph obtained by associating the average velocity on a time interval with the midpoint clock time on that interval. We find the time interval between each pair of midpoint times, and the change in average velocities between those two midpoint times. Dividing velocity change by time interval we get the rate of velocity change, or acceleration. **
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RESPONSE --> ok-- i did not write down HOW to plot velocity vs. time.....
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07:35:30 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, I missed problem 4. initial velocity = 18 cm/s think of this as horizontal velocity. this horizontal velocity can only occur as long as the object is falling at 980 cm/sec^2. ( acceleration due to gravity) so...... d=v0t + 1/at^2 d= 95 cm vo (vertical)= 0 1/2 a= 460 cm/sec^2 t= ? 95 cm = 460 cm/sec^2* t^2 95cm/460 cm/sec^2= t^2 =.207 sec^2 .45 secs= time now we can use the time to find the horizontal distance the projectile travels during that time. d= Vave * t, so .45sec*18cm/sec= so the projectile should travel 8.1 cm. why did you get .194 for 95/460? never mind! 1/2*980= 490!!! 95/490=.194 ...........
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08:01:20 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 --> vf^2 = 2ad where v0 is 0 (13.9cm/sec)^2 = 2a * 40 cm 193.2 cm^2/sec^2/80cm = 2.4 cm/sec/sec final velocity is given at 13.9 cm/sec acceleration is found to be 2.4 cm/sec^2 the time would be 13.9cm/sec= 2.4 cm/sec^2 * t or 5.8 seconds
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08:02:48 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 actually used the quadratic to find the acceleration......
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08:03:05 002.
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RESPONSE --> 002????
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"
course phy good luck! òæ×;َËÚ‰Ê|ízäñàCÎ Q¡îîâìassignment #004
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15:18:40 How do we reason out the process of determining acceleration from rest given displacement and time duration?
......!!!!!!!!...................................
RESPONSE --> If an obect starts from rest, and travels a given distance in a given amount of time, then we canfind the acceleration. first, we find the average velocity by dividing distance traveled by the time (ie, a ball rolls 15 cm in 3 sec--> 15/3= 5 cm/sec) then, knowing that the average velocity is the final velocity/2 when an object starts at rest [ (vf + v0)/2 = vave ---> (vf + 0)/2=5 cm/sec --> vf/2 = 5 cm/sec --> vf= 10cm/sec] we find the final velocity-- in this case 10 cm/sec. the acceleration is the difference velocity over time, so (vf-v0)/t in the example, (10 cm/sec - 0 cm/sec)/3 sec = 3.33 cm/sec/sec
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15:19:22 ** COMMON ERROR: By first finding the average velocity. Also we must understand this gives us the change in velocity which gives us the acceleration. INSTRUCTOR COMMENT: Average acceleration is not related to average velocity. It is related to the change in velocity, but change in velocity is not the same thing as acceleration. Acceleration is the rate of change of velocity. ANSWER TO QUESTION: Average acceleration is defined to be the average rate at which velocity changes, which is change in velocity/change in clock time, or equivalently change in velocity/time duration. However we do not know the change in velocity, nor from the given information can we directly determine the change in velocity. So we have to look at what we can reason out from what we know. Given displacement from rest and time duration we calculate average velocity: vAve = `ds / `dt = displacement/time duration. Since the initial velocity is 0, the final velocity is double the average so we find the final velocity by doubling the average velocity. Now that we know the initial velocity 0 and the final velocity we can find change in velocity by subtracting initial vel from final vel. Dividing change in velocity by a time duration we finally obtain the average acceleration. **
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RESPONSE --> exactly
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¯±ç×Ø¿}Çy䌀cÉxÅòÉ¿ËðˆPïóÇʲD± assignment # òæ×;َËÚ‰Ê|ízäñàCÎ Q¡îîâì Liberal Arts Mathematics I 08-03-2006
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21:35:29 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 --> ok
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23:24:54 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 --> ok, with ramp elevated 1.8 cm the average velocity is 78cm/2.722297 or 28.7 cm/sec associated with a slope created by a right sided triangle with height 1.8 and hypotenuse 78 cm-- 1.8^2 + width^2 = 78^2 which is 78^2-1.8^2 = w^2 = 77.97cm 1.8/77.97= a slope of .023 cm. 28.7 cm/sec slope of .023cm with ramp elevated 4.2 cm, the average velocity is 78cm/2.518101 sec or 30.97572337cm/sec ~ 31cm/sec the slope here is 78^2- 4.2^2= w^2 w=77.88 rise over run = 4.2/77.88 or .052 cm slope 30.9757cm/sec with 6.7 cm ramp, the average velocity 78cm/2.6606sec = 29.31669548 cm/sec ~ 29.3 cm/sec the slope is 78^2-6.7^2= 77.71^2 rise over run= 6.7/77.71 or a slope of .086 cm slope with a velocity of 29.3 cm/sec since the highest velocity is actually associated with a lower slope, the data does not support the hypopthesis
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23:25:43 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 -->
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23:26:18 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 --> it doesn't form a line
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23:58:58 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 --> since vave * time= distance, and since vave= (vf+v0)/2, then (6.31579 + v0)/2*(3.8)= 50cm 13.1579 cm/s= (6.31579 cm/s + v0)/2 26.3158 cm/s - 6.31579 = v0 19.9999~ 20 cm/sec acceleration is the (vf-v0)/3.8 seconds in this case there is negative acceleration -3.6011 cm/sec^2
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00:00:40 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 --> Enter, as appropriate, an answer to the question, a critique of your answer in response to a given answer, your insights regarding the situation at this point, notes to yourself, or just an OK. Always critique your solutions by describing any insights you had or errors you makde, and by explaining how you can make use of the insight or how you now know how to avoid certain errors. Also pose for the instructor any question or questions that you have related to the problem or series of problems. yes.
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07:00:24 How do we obtain an approximate velocity vs. clock time table and graph from a series of position vs. clock time observations? How do we then obtain an approximate acceleration vs. clock timetable and graph?
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RESPONSE --> If we have a graph of position vs. clock time, then the slope of the graph between any two clock times represents the average velocity between those times. by measuring the slope at short time intervals, we can then recreate (a fairly accurate) velocity vs time graph.
.................................................
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07:01:25 ** If we start with a position vs. clock time graph we do divide the graph into intervals. On each interval velocity, which is rate of change of position, is represented by the slope of the graph. So we have to calculate the slope of the graph on each interval. If we then graph the slopes vs. the midpoint times of the intervals we get a good approximation of the velocity vs. clock time graph. The velocity at a given instant is the slope of the position vs. clock time graph. STUDENT ANSWER: We first calculate the time interval and displacement between each pair of data points. We use these calculations to calculate the average velocity. To obtain an approximate accelerations. Clock timetable and graph by associating the average velocity on a time interval with the midpoint clock time on that interval. INSTRUCTOR CRITIQUE: Using your words and amplifying a bit: We first calculate the time interval and displacement between each pair of data points. We use these results to calculate the average velocity, dividing displacement by time interval for each interval. Then we make a table, showing the average velocity vs. the midpoint time for each time interval. To obtain approximate accelerations we use the table and graph obtained by associating the average velocity on a time interval with the midpoint clock time on that interval. We find the time interval between each pair of midpoint times, and the change in average velocities between those two midpoint times. Dividing velocity change by time interval we get the rate of velocity change, or acceleration. **
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RESPONSE --> ok-- i did not write down HOW to plot velocity vs. time.....
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07:35:30 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, I missed problem 4. initial velocity = 18 cm/s think of this as horizontal velocity. this horizontal velocity can only occur as long as the object is falling at 980 cm/sec^2. ( acceleration due to gravity) so...... d=v0t + 1/at^2 d= 95 cm vo (vertical)= 0 1/2 a= 460 cm/sec^2 t= ? 95 cm = 460 cm/sec^2* t^2 95cm/460 cm/sec^2= t^2 =.207 sec^2 .45 secs= time now we can use the time to find the horizontal distance the projectile travels during that time. d= Vave * t, so .45sec*18cm/sec= so the projectile should travel 8.1 cm. why did you get .194 for 95/460? never mind! 1/2*980= 490!!! 95/490=.194 ...........
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08:01:20 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 --> vf^2 = 2ad where v0 is 0 (13.9cm/sec)^2 = 2a * 40 cm 193.2 cm^2/sec^2/80cm = 2.4 cm/sec/sec final velocity is given at 13.9 cm/sec acceleration is found to be 2.4 cm/sec^2 the time would be 13.9cm/sec= 2.4 cm/sec^2 * t or 5.8 seconds
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08:02:48 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 actually used the quadratic to find the acceleration......
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08:03:05 002.
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RESPONSE --> 002????
.................................................
"
course phy good luck! òæ×;َËÚ‰Ê|ízäñàCÎ Q¡îîâìassignment #004
......!!!!!!!!...................................
15:18:40 How do we reason out the process of determining acceleration from rest given displacement and time duration?
......!!!!!!!!...................................
RESPONSE --> If an obect starts from rest, and travels a given distance in a given amount of time, then we canfind the acceleration. first, we find the average velocity by dividing distance traveled by the time (ie, a ball rolls 15 cm in 3 sec--> 15/3= 5 cm/sec) then, knowing that the average velocity is the final velocity/2 when an object starts at rest [ (vf + v0)/2 = vave ---> (vf + 0)/2=5 cm/sec --> vf/2 = 5 cm/sec --> vf= 10cm/sec] we find the final velocity-- in this case 10 cm/sec. the acceleration is the difference velocity over time, so (vf-v0)/t in the example, (10 cm/sec - 0 cm/sec)/3 sec = 3.33 cm/sec/sec
.................................................
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15:19:22 ** COMMON ERROR: By first finding the average velocity. Also we must understand this gives us the change in velocity which gives us the acceleration. INSTRUCTOR COMMENT: Average acceleration is not related to average velocity. It is related to the change in velocity, but change in velocity is not the same thing as acceleration. Acceleration is the rate of change of velocity. ANSWER TO QUESTION: Average acceleration is defined to be the average rate at which velocity changes, which is change in velocity/change in clock time, or equivalently change in velocity/time duration. However we do not know the change in velocity, nor from the given information can we directly determine the change in velocity. So we have to look at what we can reason out from what we know. Given displacement from rest and time duration we calculate average velocity: vAve = `ds / `dt = displacement/time duration. Since the initial velocity is 0, the final velocity is double the average so we find the final velocity by doubling the average velocity. Now that we know the initial velocity 0 and the final velocity we can find change in velocity by subtracting initial vel from final vel. Dividing change in velocity by a time duration we finally obtain the average acceleration. **
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RESPONSE --> exactly
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¯±ç×Ø¿}Çy䌀cÉxÅòÉ¿ËðˆPïóÇʲD± assignment # òæ×;َËÚ‰Ê|ízäñàCÎ Q¡îîâì Liberal Arts Mathematics I 08-03-2006
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21:35:29 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 --> ok
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23:24:54 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 --> ok, with ramp elevated 1.8 cm the average velocity is 78cm/2.722297 or 28.7 cm/sec associated with a slope created by a right sided triangle with height 1.8 and hypotenuse 78 cm-- 1.8^2 + width^2 = 78^2 which is 78^2-1.8^2 = w^2 = 77.97cm 1.8/77.97= a slope of .023 cm. 28.7 cm/sec slope of .023cm with ramp elevated 4.2 cm, the average velocity is 78cm/2.518101 sec or 30.97572337cm/sec ~ 31cm/sec the slope here is 78^2- 4.2^2= w^2 w=77.88 rise over run = 4.2/77.88 or .052 cm slope 30.9757cm/sec with 6.7 cm ramp, the average velocity 78cm/2.6606sec = 29.31669548 cm/sec ~ 29.3 cm/sec the slope is 78^2-6.7^2= 77.71^2 rise over run= 6.7/77.71 or a slope of .086 cm slope with a velocity of 29.3 cm/sec since the highest velocity is actually associated with a lower slope, the data does not support the hypopthesis
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23:25:43 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 -->
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23:26:18 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 --> it doesn't form a line
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23:58:58 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 --> since vave * time= distance, and since vave= (vf+v0)/2, then (6.31579 + v0)/2*(3.8)= 50cm 13.1579 cm/s= (6.31579 cm/s + v0)/2 26.3158 cm/s - 6.31579 = v0 19.9999~ 20 cm/sec acceleration is the (vf-v0)/3.8 seconds in this case there is negative acceleration -3.6011 cm/sec^2
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00:00:40 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 --> Enter, as appropriate, an answer to the question, a critique of your answer in response to a given answer, your insights regarding the situation at this point, notes to yourself, or just an OK. Always critique your solutions by describing any insights you had or errors you makde, and by explaining how you can make use of the insight or how you now know how to avoid certain errors. Also pose for the instructor any question or questions that you have related to the problem or series of problems. yes.
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07:00:24 How do we obtain an approximate velocity vs. clock time table and graph from a series of position vs. clock time observations? How do we then obtain an approximate acceleration vs. clock timetable and graph?
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RESPONSE --> If we have a graph of position vs. clock time, then the slope of the graph between any two clock times represents the average velocity between those times. by measuring the slope at short time intervals, we can then recreate (a fairly accurate) velocity vs time graph.
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07:01:25 ** If we start with a position vs. clock time graph we do divide the graph into intervals. On each interval velocity, which is rate of change of position, is represented by the slope of the graph. So we have to calculate the slope of the graph on each interval. If we then graph the slopes vs. the midpoint times of the intervals we get a good approximation of the velocity vs. clock time graph. The velocity at a given instant is the slope of the position vs. clock time graph. STUDENT ANSWER: We first calculate the time interval and displacement between each pair of data points. We use these calculations to calculate the average velocity. To obtain an approximate accelerations. Clock timetable and graph by associating the average velocity on a time interval with the midpoint clock time on that interval. INSTRUCTOR CRITIQUE: Using your words and amplifying a bit: We first calculate the time interval and displacement between each pair of data points. We use these results to calculate the average velocity, dividing displacement by time interval for each interval. Then we make a table, showing the average velocity vs. the midpoint time for each time interval. To obtain approximate accelerations we use the table and graph obtained by associating the average velocity on a time interval with the midpoint clock time on that interval. We find the time interval between each pair of midpoint times, and the change in average velocities between those two midpoint times. Dividing velocity change by time interval we get the rate of velocity change, or acceleration. **
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RESPONSE --> ok-- i did not write down HOW to plot velocity vs. time.....
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07:35:30 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, I missed problem 4. initial velocity = 18 cm/s think of this as horizontal velocity. this horizontal velocity can only occur as long as the object is falling at 980 cm/sec^2. ( acceleration due to gravity) so...... d=v0t + 1/at^2 d= 95 cm vo (vertical)= 0 1/2 a= 460 cm/sec^2 t= ? 95 cm = 460 cm/sec^2* t^2 95cm/460 cm/sec^2= t^2 =.207 sec^2 .45 secs= time now we can use the time to find the horizontal distance the projectile travels during that time. d= Vave * t, so .45sec*18cm/sec= so the projectile should travel 8.1 cm. why did you get .194 for 95/460? never mind! 1/2*980= 490!!! 95/490=.194 ...........
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08:01:20 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 --> vf^2 = 2ad where v0 is 0 (13.9cm/sec)^2 = 2a * 40 cm 193.2 cm^2/sec^2/80cm = 2.4 cm/sec/sec final velocity is given at 13.9 cm/sec acceleration is found to be 2.4 cm/sec^2 the time would be 13.9cm/sec= 2.4 cm/sec^2 * t or 5.8 seconds
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08:02:48 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 actually used the quadratic to find the acceleration......
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08:03:05 002.
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RESPONSE --> 002????
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course phy good luck! òæ×;َËÚ‰Ê|ízäñàCÎ Q¡îîâìassignment #004
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15:18:40 How do we reason out the process of determining acceleration from rest given displacement and time duration?
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RESPONSE --> If an obect starts from rest, and travels a given distance in a given amount of time, then we canfind the acceleration. first, we find the average velocity by dividing distance traveled by the time (ie, a ball rolls 15 cm in 3 sec--> 15/3= 5 cm/sec) then, knowing that the average velocity is the final velocity/2 when an object starts at rest [ (vf + v0)/2 = vave ---> (vf + 0)/2=5 cm/sec --> vf/2 = 5 cm/sec --> vf= 10cm/sec] we find the final velocity-- in this case 10 cm/sec. the acceleration is the difference velocity over time, so (vf-v0)/t in the example, (10 cm/sec - 0 cm/sec)/3 sec = 3.33 cm/sec/sec
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15:19:22 ** COMMON ERROR: By first finding the average velocity. Also we must understand this gives us the change in velocity which gives us the acceleration. INSTRUCTOR COMMENT: Average acceleration is not related to average velocity. It is related to the change in velocity, but change in velocity is not the same thing as acceleration. Acceleration is the rate of change of velocity. ANSWER TO QUESTION: Average acceleration is defined to be the average rate at which velocity changes, which is change in velocity/change in clock time, or equivalently change in velocity/time duration. However we do not know the change in velocity, nor from the given information can we directly determine the change in velocity. So we have to look at what we can reason out from what we know. Given displacement from rest and time duration we calculate average velocity: vAve = `ds / `dt = displacement/time duration. Since the initial velocity is 0, the final velocity is double the average so we find the final velocity by doubling the average velocity. Now that we know the initial velocity 0 and the final velocity we can find change in velocity by subtracting initial vel from final vel. Dividing change in velocity by a time duration we finally obtain the average acceleration. **
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RESPONSE --> exactly
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¯±ç×Ø¿}Çy䌀cÉxÅòÉ¿ËðˆPïóÇʲD± assignment # òæ×;َËÚ‰Ê|ízäñàCÎ Q¡îîâì Liberal Arts Mathematics I 08-03-2006
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21:35:29 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 --> ok
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23:24:54 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 --> ok, with ramp elevated 1.8 cm the average velocity is 78cm/2.722297 or 28.7 cm/sec associated with a slope created by a right sided triangle with height 1.8 and hypotenuse 78 cm-- 1.8^2 + width^2 = 78^2 which is 78^2-1.8^2 = w^2 = 77.97cm 1.8/77.97= a slope of .023 cm. 28.7 cm/sec slope of .023cm with ramp elevated 4.2 cm, the average velocity is 78cm/2.518101 sec or 30.97572337cm/sec ~ 31cm/sec the slope here is 78^2- 4.2^2= w^2 w=77.88 rise over run = 4.2/77.88 or .052 cm slope 30.9757cm/sec with 6.7 cm ramp, the average velocity 78cm/2.6606sec = 29.31669548 cm/sec ~ 29.3 cm/sec the slope is 78^2-6.7^2= 77.71^2 rise over run= 6.7/77.71 or a slope of .086 cm slope with a velocity of 29.3 cm/sec since the highest velocity is actually associated with a lower slope, the data does not support the hypopthesis
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23:25:43 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 -->
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23:26:18 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 --> it doesn't form a line
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23:58:58 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 --> since vave * time= distance, and since vave= (vf+v0)/2, then (6.31579 + v0)/2*(3.8)= 50cm 13.1579 cm/s= (6.31579 cm/s + v0)/2 26.3158 cm/s - 6.31579 = v0 19.9999~ 20 cm/sec acceleration is the (vf-v0)/3.8 seconds in this case there is negative acceleration -3.6011 cm/sec^2
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00:00:40 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 --> Enter, as appropriate, an answer to the question, a critique of your answer in response to a given answer, your insights regarding the situation at this point, notes to yourself, or just an OK. Always critique your solutions by describing any insights you had or errors you makde, and by explaining how you can make use of the insight or how you now know how to avoid certain errors. Also pose for the instructor any question or questions that you have related to the problem or series of problems. yes.
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07:00:24 How do we obtain an approximate velocity vs. clock time table and graph from a series of position vs. clock time observations? How do we then obtain an approximate acceleration vs. clock timetable and graph?
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RESPONSE --> If we have a graph of position vs. clock time, then the slope of the graph between any two clock times represents the average velocity between those times. by measuring the slope at short time intervals, we can then recreate (a fairly accurate) velocity vs time graph.
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07:01:25 ** If we start with a position vs. clock time graph we do divide the graph into intervals. On each interval velocity, which is rate of change of position, is represented by the slope of the graph. So we have to calculate the slope of the graph on each interval. If we then graph the slopes vs. the midpoint times of the intervals we get a good approximation of the velocity vs. clock time graph. The velocity at a given instant is the slope of the position vs. clock time graph. STUDENT ANSWER: We first calculate the time interval and displacement between each pair of data points. We use these calculations to calculate the average velocity. To obtain an approximate accelerations. Clock timetable and graph by associating the average velocity on a time interval with the midpoint clock time on that interval. INSTRUCTOR CRITIQUE: Using your words and amplifying a bit: We first calculate the time interval and displacement between each pair of data points. We use these results to calculate the average velocity, dividing displacement by time interval for each interval. Then we make a table, showing the average velocity vs. the midpoint time for each time interval. To obtain approximate accelerations we use the table and graph obtained by associating the average velocity on a time interval with the midpoint clock time on that interval. We find the time interval between each pair of midpoint times, and the change in average velocities between those two midpoint times. Dividing velocity change by time interval we get the rate of velocity change, or acceleration. **
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RESPONSE --> ok-- i did not write down HOW to plot velocity vs. time.....
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07:35:30 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, I missed problem 4. initial velocity = 18 cm/s think of this as horizontal velocity. this horizontal velocity can only occur as long as the object is falling at 980 cm/sec^2. ( acceleration due to gravity) so...... d=v0t + 1/at^2 d= 95 cm vo (vertical)= 0 1/2 a= 460 cm/sec^2 t= ? 95 cm = 460 cm/sec^2* t^2 95cm/460 cm/sec^2= t^2 =.207 sec^2 .45 secs= time now we can use the time to find the horizontal distance the projectile travels during that time. d= Vave * t, so .45sec*18cm/sec= so the projectile should travel 8.1 cm. why did you get .194 for 95/460? never mind! 1/2*980= 490!!! 95/490=.194 ...........
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08:01:20 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 --> vf^2 = 2ad where v0 is 0 (13.9cm/sec)^2 = 2a * 40 cm 193.2 cm^2/sec^2/80cm = 2.4 cm/sec/sec final velocity is given at 13.9 cm/sec acceleration is found to be 2.4 cm/sec^2 the time would be 13.9cm/sec= 2.4 cm/sec^2 * t or 5.8 seconds
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08:02:48 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 actually used the quadratic to find the acceleration......
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08:03:05 002.
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RESPONSE --> 002????
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