04-MAR-2015: Non-Constant Acceleration Problem

Purpose:  This lab is based on a non-constant acceleration problem detailed below:

     - "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 level ground.  At that point a rocket mounted 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."

The purposes of this lab include solving this problem analytically as well as numerically (using Excel).  Using the given data above we can obtain a function representing the elephant's acceleration at any given time.  

Materials:  The materials for this lab include: Microsoft Excel, integral calculus techniques.

Procedure:  Solving this problem analytically involves integrating the acceleration function with respect to time to obtain the elephant's velocity function, then integrating once again to obtain the elephant's position function.  Since the problem states to find the distance of the elephant once it is at rest, we set the derived velocity function equal to 0 and solve for t.  This gives us the time the elephant is at rest.  We then evaluate our derived position function at this time to give us the position of the elephant when it comes to rest (which is what the problem is asking for).  Relevant equations are listed below.

     - Mass of rocket over time function

       m(t) = 1500 kg - 20 kg/s*t

     - Deriving acceleration function

     a(t) = Fnet / m(t)

     a(t) = -8000 N / (6500 kg - 20 kg/s * t)

     a(t) = (-400 / (325 - t) ) m/s^2

     - Derived velocity function

     v(t) = [25 - 400ln(325)] + 400ln(325 - t)

     - Solving for time elephant comes to rest

     0 = [25 - 400ln(325)] + 400ln(325 - t)

     t = 19.70 s

     - Derived position function

     x(t) = [25 - 400ln(325)]t + 400[(t - 325)*ln(325 - t) - t + 325ln(325)]

     - Solving for position elephant comes to rest

     x(19.70) = 248.7 m

The equations above are simply a summary of the work involved in integrating this manually, the overall work would take up a significantly larger portion of this blog entry.  We were lucky the initial acceleration function was able to be integrated manually, as this is not always the case.  After solving this problem analytically, we then began to solve it numerically utilizing Excel to input and manipulate our data.  Depicted below is the beginning of said Excel spreadsheet.

(Top of Excel spreadsheet)

If we scroll down to where column A (time) is equal to our analytically calculated value of 19.70 s, we see that the value of column F (position) is in accordance with our calculated value of 248.7 m.

(Section of spreadsheet which reveals the position at 19.7 s to be 248.70 m)

Conclusions:  

1)  As it is displayed in the above screenshot, the value of the elephant's position we calculated analytically using integration (248.7 m) is equivalent to the position of the elephant in our spreadsheet's data table.

2)  You know the time interval is small enough when the velocity is extremely close to 0, (- 0.01 s in this case).  If there were no analytic result with which to compare our data table to, another way to determine the final position of the elephant is to examine the values of position in the table.  Taking a closer look reveals that there is a MAX number of 248.70 m, values for position before and after are all less than that at t = 19.70 s.