8-March-17: Lab 3 Non-Constant acceleration problem/Activity

Eric Chong

Lab 3 Non-Constant acceleration problem/Activity

The Problem: A 5000-kg elephant on frictionless roller skates is going 25 m/s when it gets to the bottom of a hill and arrives on a level ground. At that point a 1500-kg rocket moounted on the elephant's back generates a constant 8000 N thrust opposite the elephant's direction of motion. The mass of the rocket changes with time (due to burning the fuel at a rate of 20 kg/s) so that the

m(t) = 1500 kg - 20 kg/s*t. Find how far the elephant goes before coming to rest.

Introduction: For this activity, we are given a problem about an elephant with a rocket that is thrusting backwards while losing mass. There are two ways to approach this problem. The first way is the analytical approach, which essentially relies on the use of calculus, physics laws, and other equations in order to find the distance the elephant traveled. We did not do this ourselves, for our professor showed us how to do it in the handout because this approach is very time-consuming. Here is how he did it:

The other way to approach this is the numerical approach, where we use an excel spreadsheet in order to link values of certain variables with other variables in order to help us find the distance at the elephant travels. Here is the initial layout of the spreadsheet:

(Note: M0 is the initial mass, V0 is the initial velocity, b is the burn rate, F is the force of the thrust, t is the time, x is the position, and a is acceleration. The rest of the variables, such as v_avg and delta x should be self-explanatory.)

We first filled out the given values in the spreadsheet (M0, v0, b, F, and delta t). Then, we manipulated the columns at the bottom portion of the spreadsheet by filling out formulas that correspond to them.

For t (time), the formula is equal to the previous time plus delta t, with the initial time being 0. This gives the next subsequent values of time based on the change in time.

For a (acceleration), the formula is the force divided by the quantity of initial mass minus b (burn rate) times t (time).

For a_avg (average acceleration), the formula is the final acceleration plus the initial acceleration of the time interval, all over delta t.

For delta v (change in velocity), the formula is the average acceleration at that time interval multiplied by the change in time.

For v (velocity), the formula is the previous velocity plus delta v.

For v_avg (average velocity), the formula is the final velocity plus the initial velocity of the time interval, all over change in time.

For delta x (change in position), the formula is the average velocity at that time interval times the change in time.

And lastly for x (position), the formula is the previous position plus the change in position.

After we filled out the formulas and inputted the values and used the fill down function, the resulted spreadsheet looks something like this:

We first set the delta t as 1 second and recorded the values. We then set delta t as 0.1 seconds and 0.05 seconds and recorded their data. Here are the values in the form of a table:

We observed that the smaller the delta t, the more accurate the position is, since essentially we are making thinner rectangles under the position vs. time curve and minimizing the area left out by the corners of the rectangles, the same idea as taking the integral, when the rectangles are infinitely thin under the curve. As expected, through this method, the answer is around 248.7 meters, the same distance as when solved through the analytical approach.

Conclusion: Overall, this little exercise is a means to show how each value relates to one another in the form of an excel spreadsheet. It helps us practice the fundamentals of deriving equations and thinking how we can obtain the values of each variable.

1.) The answers are relatively close, as through the analytical approach the answer is 248 m, and through the numerical approach the answer is around 248.7 m.

2.) We can figure out that the time interval is small enough and find the answer, even without the reference of the analytical result, by looking at the digits of the results based on the value of the time interval (1 s, 0.1 s, or 0.05 s), and seeing at which point does the value stop changing at a significant rate. Once we notice that the results stops changing significantly for a specific value of the time interval, we can estimate to the decimal point that matters to us, and in this case is to the tenths place (248.7 m), and we can then claim that the time interval is small enough. For example, in the previous picture, we see that the time interval of 1 second is too large, as the values for the distance varies too much within that time interval. Thus, by reducing the time interval to 0.1 second and then 0.05 second, we found the result to be more accurate. That is how we knew when the time interval is too large, and when it is small enough.

3.) For this problem, we would do the same numerical approach with the spreadsheet. After setting everything up and getting the values, here are the results:

According to the data table from the numerical approach, the distance the elephant would travel is around 164.04 m (as seen in the highlighted cells).