course Phy 121
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08:38:47 `q001. The accepted value of the acceleration of gravity is approximately 980 cm/s^2 or 9.8 m/s^2. This will be the acceleration, accurate at most places within 1 cm/s^2, of any object which falls freely, that is without the interference of any other force, near the surface of the Earth. If you were to step off of a table and were to fall 1 meter without hitting anything, you would very nearly approximate a freely falling object. How fast would you be traveling when you reached the ground?
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RESPONSE --> Use the equation vf^2 = v0^2 + 2 a `ds to find vf. Take the square root of vf to get vf by itself. vf = +- 'sqrt (v0^2 + 2 a 'ds) vf = +- `sqrt ( 0^2 + 2 * 9.8 m/s^2 * 1 m) vf = +- `sqrt ( 19.6 m^2 / s^2) vf = +- 4.4 m/s
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08:39:05 02-10-2006 08:39:05 You would have an initial vertical velocity of 0, and would accelerate at 9.8 m/s^2 in the same direction as your 1 meter vertical displacement. You would also have a slight horizontal velocity (you don't step off of a table without moving a bit in the horizontal direction, and you would very likely maintain a small horizontal velocity as you fell), but this would have no effect on your vertical motion. So your vertical velocity is a uniform acceleration with v0 = 0, `ds = 1 meter and a = 9.8 m/s^2. The equation vf^2 = v0^2 + 2 a `ds contains the three known variables and can therefore be used to find the desired final velocity. We obtain vf = +- `sqrt( v0^2 + 2 a `ds) = +- `sqrt ( 0^2 + 2 * 9.8 m/s^2 * 1 m)= +- `sqrt ( 19.6 m^2 / s^2) = +- 4.4 m/s, approx. Since the acceleration and displacement were in the direction chosen as positive, we conclude that the final velocity will be in the same direction and we choose the solution vf = +4.4 m/s.
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NOTES -------> Since final velocity is assumed to be in the same direction.
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08:48:27 `q002. If you jump vertically upward, leaving the ground with a vertical velocity of 3 m/s, how high will you be at the highest point of your jump? Note that as soon as you leave the ground, you are under the influence of only the gravitational force. All the forces that you exerted with your legs and other parts of your body to attain the 3 m/s velocity have done their work and are no longer acting on you. All you have to show for it is that 3 m/s velocity. So as soon as you leave the ground, you begin experiencing an acceleration of 9.8 m/s^2 in the downward direction. Now again, how high will you be at the highest point of your jump?
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RESPONSE --> There is not a uniform equation with all these variables in them so we have to peice it together. First find the time using the following equation: `dt = (vf - v0) / a 'dt = (0 - 3 m/s) / (-9.8 m/s^2) 'dt = .3 sec Then use all the information to find the 'ds `ds = (vf + v0) / 2 * `dt 'ds = (3 m/s + 0 m/s) / 2 * .3 sec 'ds = .45 m
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08:48:42 From the instant the leave the ground until the instant you reach your highest point, you have an acceleration of 9.8 m/s^2 in the downward direction. Since you are jumping upward, and since we can take our choice of whether upward or downward is the positive direction, we choose the upward direction as positive. You might have chosen the downward direction, and we will see in a moment how you should have proceeded after doing so. For now, using the upward direction as positive, we see that you have an initial velocity of v0 = + 3 m/s and an acceleration of a = -9.8 m/s^2. In order to use any of the equations of motion, each of which involves four variables, you should have the values of three variables. So far you only have two, v0 and a. {}What other variable might you know? If you think about it, you will notice that when objects tossed in the air reach their highest point they stop for an instant before falling back down. That is precisely what will happen to you. At the highest point your velocity will be 0. Since the highest point is the last point we are considering, we see that for your motion from the ground to the highest point, vf = 0. Therefore we are modeling a uniform acceleration situation with v0 = +3 m/s, a = -9.8 m/s^2 and vf = 0. We wish to find the displacement `ds. Unfortunately none of the equations of uniformly accelerated motion contain the four variables v0, a, vf and `ds. This situation can be easily reasoned out from an understanding of the basic quantities. We can find the change in velocity to be -3 meters/second; since the acceleration is equal to the change in velocity divided by the time interval we quickly determine that the time interval is equal to the change in velocity divided by the acceleration, which is `dt = -3 m/s / (-9.8 m/s^2) = .3 sec, approx.; then we multiply the .3 second time interval by the 1.5 m/s average velocity to obtain `ds = .45 meters. However if we wish to use the equations, we can begin with the equation vf = v0 + a `dt and solve to find `dt = (vf - v0) / a = (0 - 3 m/s) / (-9.8 m/s^2) = .3 sec. We can then use the equation `ds = (vf + v0) / 2 * `dt = (3 m/s + 0 m/s) / 2 * .3 sec = .45 m. This solution closely parallels and is completely equivalent to the direct reasoning process, and shows that and initial velocity of 3 meters/second should carry a jumper to a vertical height of .45 meters, approximately 18 inches. This is a fairly average vertical jump. If the negative direction had been chosen as positive then we would have a = +9.8 m/s^2, v0 = -3 m/s^2 (v0 is be in the direction opposite the acceleration so if acceleration is positive then initial velocity is negative) and again vf = 0 m/s (0 m/s is the same whether going up or down). The steps of the solution will be the same and the same result will be obtained, except that `ds will be -.45 m--a negative displacement, but where the positive direction is down. That is we move .45 m in the direction opposite to positive, meaning we move .45 meters upward.
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RESPONSE --> ok
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10:01:21 `q003. If you roll a ball along a horizontal table so that it rolls off the edge of the table at a velocity of 3 m/s, the ball will continue traveling in the horizontal direction without changing its velocity appreciably, and at the same time will fall to the floor in the same time as it would had it been simply dropped from the edge of the table. If the vertical distance from the edge of the table to the floor is .9 meters, then how far will the ball travel in the horizontal direction as it falls?
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RESPONSE --> Use the equation `ds = v0 `dt + .5 a `dt^2 And solve for 'dt 'dt = `sqrt( 2 `ds / a) 'dt = `sqrt(2 * .9 m / (9.8 m/s^2) ) 'dt = .42 sec Now multiply the displacement by the change in time to find how far the ball fell: 3m/s * .42 s = 1.26 m
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10:02:33 02-10-2006 10:02:33 A ball dropped from rest at a height of .9 meters will fall to the ground with a uniform vertical acceleration of 9.8 m/s^2 downward. Selecting the downward direction as positive we have `ds = .9 meters, a = 9.8 m/s^2 and v0 = 0. Using the equation `ds = v0 `dt + .5 a `dt^2 we see that v0 = 0 simplifies the equation to `ds = .5 a `dt^2, so `dt = `sqrt( 2 `ds / a) = `sqrt(2 * .9 m / (9.8 m/s^2) ) = .42 sec, approx.. Since the ball rolls off the edge of the table with only a horizontal velocity, its initial vertical velocity is still zero and it still falls to the floor in .42 seconds. Since its horizontal velocity remains at 3 m/s, it travels through a displacement of 3 m/s * .42 sec = 1.26 meters in this time.
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NOTES -------> I state the last part of the answer wrong the velocity times the change in time gives you displacement.
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瀱khbC~Ÿŭ assignment #008 [뺗܃ܝ Physics I Class Notes 02-10-2006
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10:08:36 What is the meaning of the x intercept of the graph of force vs. slope?
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RESPONSE --> The x axis represents slope.
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10:10:02 02-10-2006 10:10:02 ** On this graph the x axis represents slope, the y axis represents acceleration. The x axis point occurs when y = 0, so this point represents a slope for which acceleration is zero. Assuming that 'down the incline' is the positive direction, everywhere to the right of this point the acceleration is positive and everywhere to the left is negative. The x intercept therefore represents the maximum slope for which friction is great enough to prevent the ball from accelerating down the slope. Since friction is resisting the weight component down the incline, the x intercept represents the slope when friction and the weight component tending to accelerate the cart down the slope are exactly in balance, equal and opposite. **
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NOTES -------> The x intercept represents the maximum slope for which friction is great enough to prevent the ball from accelerating down the slope.
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10:12:40 When we obtain a linear relationship between force and acceleration, is it plausible that the constant term in the equation is, within experimental uncertainty, zero?
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RESPONSE --> I suppose it is plausible. The only way that there is a zero is if it passes through the origin of the graph so there would be no acceleration or friction at that point.
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10:12:51 ** The constant term in the equation will be zero only if the graph passes through the origin. This would imply that any nonzero force will produce a nonzero acceleration, and that for example there is either no friction present or friction has been compensated for in some way. **
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RESPONSE --> ok
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10:16:43 For the force vs. acceleration relationship for the car, why should we expect that the acceleration corresponding to a net force equal to the car's weight is the acceleration of gravity?
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RESPONSE --> Since the car is free falling the force acting on it is the weight of the and gravity so this is the same as the acceleration of gravity.
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10:16:47 ** When the car is dropped and allowed to fall freely the net force acting on it is its weight, which is the force exerted on its mass by gravity. So the acceleration will be that of gravity. **
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RESPONSE --> ok
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ň㒉Pr assignment #009 [뺗܃ܝ Physics I Class Notes 02-10-2006
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10:20:03 Do the data seem to indicate, within the limits of the errors inherent in the experiment, that acceleration is the same on a constant incline regardless of where the cart is or how fast it is going?
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RESPONSE --> For the most part within the limits the acceleration appears to be constant.
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10:20:09 ** The usual results do show that the acceleration is the same, within about 10%, for any section of the ramp. Since different sections can have different lengths, which are associated with different average and final velocities, this indicates that within these limits the acceleration does appear to be constant. **
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RESPONSE --> ok
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10:20:54 Where does the force that accelerates the cart come from?
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RESPONSE --> The force is from gravity.
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10:21:53 02-10-2006 10:21:53 ** The force is the result of the gravitational force exerted on the cart by the Earth. On an incline a component of this force acts in the direction down the incline. The greater the incline the greater the proportion of the total gravitational force that acts down the incline. **
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NOTES -------> The greater the incline the greater the proportion of the force of gravity.
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10:23:40 List all the forces that act on the cart, and discuss how they affect its motion.
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RESPONSE --> Gravity pulls down. Thus the greater the slope the greater the gravity. Friction opposes motion.
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10:24:21 02-10-2006 10:24:21 ** gravity pulls straight down, but the ramp isn't able to push straight up. That leaves a component of the gravitational force parallel to the ramp; the greater the slope the greater this component. Friction opposes motion. If there is a force holding the car back, then it needs to be included in the list, as it is in your list. **
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NOTES -------> The gravitational force is parallel to the ramp. Also list any forces that might be holding the car back.
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10:31:36 Explain why a `ds is equal to the change in 1/2 v^2.
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RESPONSE --> If you look at the 4th equation of motion you see that vf^2 = v0^2 + 2 a `ds which can be rearranged to a `ds = (vf^2 - v0^2) / 2. Since vf^2 - v0^2 is the change in v^2, the a `ds is half the change in v^2.
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10:31:40 ** By the fourth equation of motion vf^2 = v0^2 + 2 a `ds. Thus a `ds = (vf^2 - v0^2) / 2. Since vf^2 - v0^2 is the change in v^2, it follows that a `ds is half the change in v^2. So a `ds is proportional to v^2. **
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RESPONSE --> ok
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10:38:16 Explain in detail why we expect Fnet `ds to be proportional to the change in v^2.
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RESPONSE --> Fnet * m = a and a 'ds = ' v^2 so fnet * 'ds = m * a 'ds
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10:38:21 ** a `ds is proportional to the change in v^2 (see the preceding question), and Fnet * m = a. So the change in Fnet * `ds is the change in m a * `ds, which for a given object is proportional to the change in v^2. **
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RESPONSE --> ok
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d؉֡ܟʴa⋓ò assignment #005 [뺗܃ܝ Physics I Vid Clips 02-11-2006
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09:53:42 Physics video clip 11 `ds on single time interval: ave. ht * width = vAve * `dt = `ds How do we calculate the displacement during a single specific time interval, given a linear v vs. t graph?
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RESPONSE --> ((v1 + v2) / 2) * (t2 - t1)
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09:54:05 02-11-2006 09:54:05 ** STUDENT SOLUTION: ((v1+v2)/2)*(t2-t1) or area of the trapezoid created by the given time interval **
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NOTES -------> or the area of a trapezoid created by the time interval
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09:54:48 If we know the initial and final velocities over some time interval, and if the rate at which velocity changes is constant, then how do we calculate the displacement of the object during that time interval?
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RESPONSE --> ((vf + v0) / 2) * 'dt
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09:55:07 02-11-2006 09:55:07 ** ((vf + v0)/2) * `dt or ((vf + v0)/2) * (t2 - t1) **
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NOTES -------> or ((vf + v0)/2) * (t2 - t1)
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09:56:40 In how many different ways can we represent the calculation of the displacement over a constant-acceleration time interval?
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RESPONSE --> Two ways, the area under the graph of v vs. t and with the equation ((vf - v0) / 2) * 'dt
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09:56:44 ** At least two: graphically as the area under the v vs. t graph and algebraically as ((vf + v0)/2) * (t2 - t1) **
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RESPONSE --> ok
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09:58:09 Physics video clip 12 continuing ph11 How does the graphical calculation connect with our common sense about velocity, displacement and time?
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RESPONSE --> The graph shows that they are all related. Graph two of the values on the x,y plane to solve for the other.
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09:58:33 02-11-2006 09:58:33 ** STUDENT SOLUTION: First of all, the graph illustrates the fact that the three are related. We can graph for two of the values on the x-y plane, in order to solve for the third one. We usually graph for velocity vs. time, to find the displacement. } STUDENT QUESTION: Is it possible to find the time by setting up a graph of distance vs. velocity or velocity vs. distance? ANSWER (UNIVERSITY PHYSICS LEVEL): Excellent question. Basically if you know the velocity at every position you have completely defined the motion; however depending on the situation it might be that the best you can do is an approximation to find x vs. t (and therefore v vs. t, again in approximation). If you know the velocity at a certain position then you can predict how long it will take to get to a nearby position; knowing the velocity at that position you can predict how long it will take to get to a new nearby position. These predictions would be approximations, the accuracy of which would depend on the magnitude of the second derivative of the velocity vs. position function. They would effectively give you an estimate of position vs. clock time, from which you could also estimate velocity vs. clock time. The situation would be that you would be able to calculate for a given position the value of the rate of position change, which is the derivative of the position. If the position function is x(t) then a graph of v vs. x would be a graph of dx/dt vs. x, so that for every x you could find dx/dt. So what you would effectively have is the differential equation dx / dt = v(x). The solution of the differential equation would be your position function x(t). Depending of v(x) there might or might not be a closed-form solution of the differential equation. If not the type of numerical approximation in the paragraph before last (which is effectively Euler's Method; there are other more sophisticated and more accurate methods of approximaton) would be your only resort, and this could be done graphically. If there is a closed-form solution to the diff eq then you might or might not be able to solve the equation x = x(t) for t in terms of x, but in any case t would be given implicitly as a function of x. **
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NOTES -------> Usually graph v vs. t to get displacement.
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Mzᆋ~ف滷z assignment #009 [뺗܃ܝ Physics I Class Notes 02-10-2006 |oɁ assignment #008 [뺗܃ܝ Physics I 02-10-2006
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10:45:14 QUESTION FROM STUDENT--Please define the differnece between Fnet and Force. See if you can answer this question.
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RESPONSE --> Fnet is simply the sum of all the forces.
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10:45:24 ** 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
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10:52:35 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 --> Just like in the class notes vf^2 = v0^2 + 2 a `ds and rearranges to a `ds = 1/2 (vf^2 - v0^2) And a `ds = k and Fnet = m a and Fnet `ds = k * ' v^2
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10:52:47 ** 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
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10:53:39 How do our experimental results confirm or refute this hypothesis?
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RESPONSE --> I am really not sure how to answer this question.
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10:55:31 02-10-2006 10:55:31 ** 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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NOTES -------> On a ramp with fixed slope a `ds is proportional to `ds. When we measured `dt for different distances from rest down the same ramp, we got the average and final velocities. The change in v^2 for each timing would be from 0 to vf^2 or 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. This would confirm the hypothesis.
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