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course phy 201
9/5 1 am
A ball rolls down a 60 cm incline and off the end. As it drops to the floor it travels an additional 16 cm in the horizontal direction, and it is known that the drop to the floor requires 0.4 seconds. From this we conclude that the ball was moving at 40 cm/s at the end of the ramp.`q001. If the ball started from rest, and if the acceleration on the ramp was uniform, then what does the graph of velocity vs. clock time look like?
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It is rising to the right, straight line. Initial is 0 cm/sec and final is 40 cm/sec.
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What new quantity or quantities can we determine, based on our graph, and what is the value of each? (Quantities in which we might be interested include acceleration, average velocity, time interval, change in velocity and possibly others; which ones can be determined directly from the graph?).
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average velocity is 20 cm/sec...time interval is 3 sec... slope or change in velocity is 13 1/3 cm/sec^2
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`q002. From the given information we know that the average velocity of the ball is 20 cm/s. Having found this, how can we determine the time interval for the motion down the ramp?
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Since we know that the average velocity is 20 cm/sec and it traveled 60 cm. The answer would be 60cm/ 20cm/sec =3 sec
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`q003. If for an interval on which the graph of v vs. t is a straight line we know two of the three quantities initial velocity, final velocity and average velocity, then we can find the third.
Suppose for such an interval we know that the initial velocity is 10 cm / sec and the final velocity is 30 cm/s. What is the average velocity?
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20 cm/sec
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Suppose we know for a different interval that the initial velocity is 20 cm/s and the average velocity is 30 cm/s. What is the final velocity?
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40 cm/sec
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On such an interval where the final velocity is 50 cm/s and the average velocity is 30 cm/s, what is the initial velocity?
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10 cm/sec
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Describe in words how you would get the final velocity for an interval on which the initial and average velocities are known.
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Take the average velocity and multiply it by 2. Then subtract the initial velocity. The result should be the final velocity.
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`q004. The average rate of change of A with respect to B is defined to be the change in A, divided by the change in B, where A and B represent specific quantities.
If the average velocity of an object on some interval is defined to be the average rate of change of position with respect to clock time on that interval, how then do we apply the definition of average rate of change to find the average velocity?
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change of position/ change of time.......
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If the average acceleration of an object on some interval is defined to be the average rate of change of velocity with respect to clock time on that interval, how then do we apply the definition of average rate of change to find the average velocity?
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rise/run= slope....change in velocity/ change in time
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What quantity could you find for an interval on which the position of an object changes by 40 cm while the clock time changes by 8 seconds?
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Average velocity 40cm/8sec = 5cm/sec final velocity 2*5cm/sec = 10 cm/sec
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The final velocity would be valid if we had the additional knowledge that initial velocity was zero and acceleration was uniform. Without those conditions you wouldn't be able to reason out the final velocity in this manner.
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What quantity could you find for an interval on which the velocity of an object changes by 300 cm/sec while the clock time changes by 20 seconds?
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Slope or change in velocity 300cm/sec / 20 sec = 15 cm/sec^2
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If you know the average rate of change of A with respect to B, and you also know the change in B, how would you find the change in A?
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Take the average and multiply it by the change in B. The answer should be the change in A.
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If you know the average rate of change of A with respect to B, and you also know the change in A, how would you find the change in B?
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Take the change in A and divide it by the average.
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If you know the average velocity of an object on an interval, and know the change in position, how do you find the change in clock time?
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Change in position divided by the average velocity.
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If you know the average acceleration of an object on an interval, and know the change in clock time, what other quantity could you find?
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Could find the change in velocity....avg. acceleration multiplied by the change in time.
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`q005. Let's return to situation we started with, the ball on the ramp. We know its initial velocity to be 0 and its final velocity to be 40 cm/s, and the length of the ramp to be 60 cm.
Summarize everything we can reason out from this data, assuming our v vs. t graph to be a straight line and using the definitions of average velocity and average acceleration.
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Average acceleration can be found..20cm/sec / 1.5sec = 13 1/3cm/sec^2........And average velocity is 60cm/ 20cm/sec = 3 sec
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`q006. For your first short rubber band, what was the color of this rubber band, and what lengths of this rubber band corresponded to what lengths of the rubber band chain?
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The rubberband was yellow......(12.5, 49.3) (19.2, 76.7) (21.1, 83.5) (17.2, 66.7)
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Give the same information for your second short rubber band.
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Rubberband was Red......(10, 47) (12.3, 61.1) (14.5, 75.6) (15.4, 81.1)
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Sketch a graph of the length of the first short rubber band vs. the length of the chain. Note the convention that a graph of y vs. x has the y quantity on the vertical axis and the x quantity on the horizontal.
Do your points lie pretty close to a single straight line? Do you think there's a tendency of your data points to curve upward or downward?
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Yes, it looks like a straight line. Curve upward as the rubberband is stretched further.
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Sketch the straight line you think lies closest, on the average, to the points of your graph. What is the slope of this line?
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rising to the right. slope 8.6cm/34.2 cm
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Repeat for your second short rubber band.
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rising to the right slope 5.4cm/34.1
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If your two short rubber bands were allowed to oppose one another, you would be able to take data for the length of one vs. the length of the other. Based on your graphs, what do you think would be the slope of a graph of the length of your first rubber band vs. the length of the second?
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The length of my first rubberband would increase at a faster rate than my second rubberband.
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Based on your two graphs, construct a graph of the length of the first short rubber band vs. the length of the second. Describe your graph.
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The slope of my second rubberband is not increasing as fast as my first rubberband.
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`q007. Give your data for the the number of dominoes vs. the length of the rubber band chain.
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Dominoes/ length (0, 31.25) (2, 33.8) (4, 36.75) (6, 41.5) (8, 50) (10,59.7)
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Sketch a graph of the number of dominoes vs. the length of the rubber band chain. Describe your graph.
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It looks like it is an increasing curve to the right.
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Sketch the straight line you think comes closest, on the average, to your data points. What is its slope?
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around 1/15cm/dom
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It looks to me as though the slope would probably be around 3 or 4 cm per domino.
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Relabel your graph using the following assumptions: Each cm marking on the ruler you used corresponds to an actual metric measurement of 0.006 meters, and each domino has a weight of 0.14 kilogram meters / second^2 (University Physics group) or 0.18 kilogram meters / second^2 (General College Physics group). In terms of your relabeling, what now is the slope of your straight line?
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1.08/.12
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`q008. Give your data for the number of cycles in 10 seconds vs. the number of dominoes.
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dominoes and cycles (4, 21) (6, 17) (8, 13) (10, 10.5)
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Sketch a graph of your data. Is it plausible that your graph could be fit by a straight line?
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No, from looking at my graph, I don't think that it is plausible to fit it to a straight line.
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Figure out the time required for a single cycles, for each number of dominoes. Give your results.
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one cycle for 4 dominoes is .48 sec...6 dominoes is .59 sec....8 dominoes is .76 sec.....10 dominoes is .93 sec
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Graph the square of the time required for a single cycle vs. the number of dominoes. Is it plausible that your graph could be well fit by a straight line? If so, give its slope.
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My graph looks like a straight line. Increasing to the right. slope is 6.67
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&#Good work. See my notes and let me know if you have questions. &#