The purpose of this lab was to solve a non-constant acceleration problem numerically using Excel. Solving the problem numerically was much faster and easier than solving it analytically.
Apparatus:
Picture 1: The non-constant acceleration problem solved in this activity
The apparatus of this lab consisted solely of a computer with the Excel program and the problem shown in the picture above. The information given from the problem is the initial velocity of the elephant (25 m/s), the final velocity of the elephant (0 m/s), the force going in the opposite direction of velocity caused by the thrust of the rocket (-8000 N), and the mass function that models the change in mass of the rocket strapped onto the elephant: m(t) = 1500 kg - (20 kg/s) t.
Abstract:
In order to solve this problem analytically (by hand), the mass function that models the whole system being studied must be found. In this example, the whole system is the elephant as well as the rocket. The mass function of the rocket is given explicitly but the elephant's is not. Conceptually, though, it is known that the mass of the elephant does not change; therefore, the mass function of the elephant is just: m(t) = 5000 kg. Now that we have both the mass function of the rocket and elephant, we can find the mass function of the whole system by just adding the two to get: m(t) = 6500 kg - (20 kg/s) t.
Now that the mass function is known, we can find the acceleration function of the system by looking at Newton's Second Law of Motion, F = ma. For our purposes, we can rearrange the formula to find acceleration (a = F/m). Using the constant force of 8000 N and the mass function, we find that for our system: a(t) = [(-8000 N) / (6500 kg - (20 kg/s) t)].
Integrating the acceleration function from 0 to t we get our change in velocity (delta v). Our velocity function is then found by adding the initial velocity to the change in velocity, as seen in this equation: v(t) = Vo + delta v. Setting the velocity function equal to zero, the time it takes for the elephant to come to rest can be found. The time was calculated to be close to 19.69 seconds.
Lastly, integrating the velocity function gives us the change in position. The position function can then be found by adding the initial position (which we take to be the origin at the bottom of the hill) and the change in position (x(t) = Xo + delta x). The final position of the elephant can then be found by plugging in the previously found value of time in the position function. The answer calculate should be about 248.7 m.
As may be seen, the integration required in this problem is complex and requires many techniques that are time consuming and increase the chance for error. Therefore, this problem was solved numerically using Excel.
Procedure:
Picture 2: Table of model for non-constant acceleration problem
Column A in a new Excel spreadsheet was labeled t for time. The change in time for each row was 0.01 s. Columm B was labeled a for acceleration. Column C was labeled a_avg for average acceleration. Column D was labeled "delta" v for change in velocity. Column E was labeled v for velocity. Column F was labeled v_avg for average velocity. Column G was labeled "delta" x for change in position. Lastly, column H was labeled x for position with respect to the origin.
Picture 3: Diagram for finding the values for each column in the Excel spreadsheet
The acceleration in column B was found by typing in the acceleration function previously found (-400/(325-t)) and plugging in the value for time at that instant (i.e. in that row). Average acceleration in column C was then found by taking the average of the acceleration value in that row as well as the row above. So for example, the average acceleration in cell C4 was found by taking the average of the acceleration values in cells B3 and B4. Next, the change in velocity in column D was found by multiplying the average acceleration by the change in time, which was 0.01 s each time. Cell E3 was value 25 since the initial velocity was 25 m/s. The rest of the values for velocity in column E were calculated by adding the change in velocity of that row with the velocity in the previous row. So, for example, the value in cell E4 was found by adding the values in cells D4 and E3. The velocity decreases as time increases because the system is decelerating (the rocket thrust is in the opposite direction of motion). The average velocity in column F was found by taking the average of the velocity at that instant as well as the velocity from the previous instant (just like how the average acceleration was found). The change in position in column G was then found by multiplying the average velocity with the change in time, which is 0.01 s. Lastly, position in column H was found by adding the change in position at that instant with the previous position (the same way it was done for velocity in column E).
Discussion:
As is known, the change in velocity is the integral of acceleration and the change in position is the integral of velocity. In addition, integration provides you with the area under a portion of the curve of a graph. The size of the portion depends on the limits of integration; a larger limit interval gives a larger area under the curve and vise versa. This was the underlying concept of the solution to this problem analytically.
However, integration cannot be done in Excel. Therefore, another way of finding velocity for acceleration and position from velocity is by manually finding the areas under the graph curves of acceleration and velocity. This method works perfectly when the curves have a constant slope i.e. acceleration or change in velocity is constant. However, problems are faced when the acceleration is not constant i.e. the slopes of the curves are not constant, as in this problem. This is because it is difficult to calculate the area of irregular shapes under the curve.
However, a common solution to this problem is by looking at very small intervals on the graph. When the interval are "small enough", the curve of the graph being studied in the interval actually becomes straight. This can be seen when you zoom in very closely at a curve in a graph, as seen below:
Picture 4: graph of y = x^2 zoomed in very close
This would take too long to be done by hand, which is integration is the best solution by hand. However, with a program like Excel, it is possible to look at very small intervals of the curve of a graph in a reasonable amount of time. With the intervals being small enough, the area under the curve of the graph can be calculate by looking at the average of the points that enclose the interval. This cannot be done with a large interval because the average is not an accurate representation of the slope of the graph. However, with very small intervals, the average becomes a very accurate representation of the slope of the curve in that interval, and the area under the curve can be found much more accurately and with much smaller error. This is the underlying reason for choosing a very small change in time for the Excel spreadsheet. This is proven when we look at the value obtained for the position of the system at 19.69 seconds highlighted below:
Picture 5: The change in position and position from the origin of the system at 19.69 seconds
The value found for the position at 19.69 seconds (248.698165 m) using the area under the curve method is very accurate and almost equal to the value found via the integration technique (248.698166 m). This proves that both techniques are equally useful when the restrictions are met. So, for example, when integration is possible and when the intervals of the curve being looked at are "small enough" such that the line connecting the limits of the interval is a straight line, respectively.
Conclusion:
The only uncertainty in this lab was the inaccuracy of the method of using the average to find the area under the curve manually. However, with smaller intervals the uncertainty becomes smaller, and with "small enough" intervals the uncertainty becomes so small that it is considered negligible.
As was explained previously, the value found by solving the problem numerically was very close to the value of solving the problem analytically. With significant figures, there would be no difference between the two and they would be considered equal.
It can be known when the interval chosen is "small enough" by comparing the numerical value to the analytical value. If the analytical value was not known such that it cannot be compared with the numerical value, it can still be known when the interval chosen is small enough by looking at the change in position. If it is so small or so close to zero such that it can be neglected, as in our problem, then we know that we are close enough and the interval is small enough.
Column A in a new Excel spreadsheet was labeled t for time. The change in time for each row was 0.01 s. Columm B was labeled a for acceleration. Column C was labeled a_avg for average acceleration. Column D was labeled "delta" v for change in velocity. Column E was labeled v for velocity. Column F was labeled v_avg for average velocity. Column G was labeled "delta" x for change in position. Lastly, column H was labeled x for position with respect to the origin.
The acceleration in column B was found by typing in the acceleration function previously found (-400/(325-t)) and plugging in the value for time at that instant (i.e. in that row). Average acceleration in column C was then found by taking the average of the acceleration value in that row as well as the row above. So for example, the average acceleration in cell C4 was found by taking the average of the acceleration values in cells B3 and B4. Next, the change in velocity in column D was found by multiplying the average acceleration by the change in time, which was 0.01 s each time. Cell E3 was value 25 since the initial velocity was 25 m/s. The rest of the values for velocity in column E were calculated by adding the change in velocity of that row with the velocity in the previous row. So, for example, the value in cell E4 was found by adding the values in cells D4 and E3. The velocity decreases as time increases because the system is decelerating (the rocket thrust is in the opposite direction of motion). The average velocity in column F was found by taking the average of the velocity at that instant as well as the velocity from the previous instant (just like how the average acceleration was found). The change in position in column G was then found by multiplying the average velocity with the change in time, which is 0.01 s. Lastly, position in column H was found by adding the change in position at that instant with the previous position (the same way it was done for velocity in column E).
Discussion:
As is known, the change in velocity is the integral of acceleration and the change in position is the integral of velocity. In addition, integration provides you with the area under a portion of the curve of a graph. The size of the portion depends on the limits of integration; a larger limit interval gives a larger area under the curve and vise versa. This was the underlying concept of the solution to this problem analytically.
However, integration cannot be done in Excel. Therefore, another way of finding velocity for acceleration and position from velocity is by manually finding the areas under the graph curves of acceleration and velocity. This method works perfectly when the curves have a constant slope i.e. acceleration or change in velocity is constant. However, problems are faced when the acceleration is not constant i.e. the slopes of the curves are not constant, as in this problem. This is because it is difficult to calculate the area of irregular shapes under the curve.
However, a common solution to this problem is by looking at very small intervals on the graph. When the interval are "small enough", the curve of the graph being studied in the interval actually becomes straight. This can be seen when you zoom in very closely at a curve in a graph, as seen below:
This would take too long to be done by hand, which is integration is the best solution by hand. However, with a program like Excel, it is possible to look at very small intervals of the curve of a graph in a reasonable amount of time. With the intervals being small enough, the area under the curve of the graph can be calculate by looking at the average of the points that enclose the interval. This cannot be done with a large interval because the average is not an accurate representation of the slope of the graph. However, with very small intervals, the average becomes a very accurate representation of the slope of the curve in that interval, and the area under the curve can be found much more accurately and with much smaller error. This is the underlying reason for choosing a very small change in time for the Excel spreadsheet. This is proven when we look at the value obtained for the position of the system at 19.69 seconds highlighted below:
The value found for the position at 19.69 seconds (248.698165 m) using the area under the curve method is very accurate and almost equal to the value found via the integration technique (248.698166 m). This proves that both techniques are equally useful when the restrictions are met. So, for example, when integration is possible and when the intervals of the curve being looked at are "small enough" such that the line connecting the limits of the interval is a straight line, respectively.
Conclusion:
The only uncertainty in this lab was the inaccuracy of the method of using the average to find the area under the curve manually. However, with smaller intervals the uncertainty becomes smaller, and with "small enough" intervals the uncertainty becomes so small that it is considered negligible.
As was explained previously, the value found by solving the problem numerically was very close to the value of solving the problem analytically. With significant figures, there would be no difference between the two and they would be considered equal.
It can be known when the interval chosen is "small enough" by comparing the numerical value to the analytical value. If the analytical value was not known such that it cannot be compared with the numerical value, it can still be known when the interval chosen is small enough by looking at the change in position. If it is so small or so close to zero such that it can be neglected, as in our problem, then we know that we are close enough and the interval is small enough.
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