4 Sept 2014: Non Constant Acceleration Lab 3

Purpose: The purpose of this lab was to solve a problem with the facilities provided by programs such as excel on the computer. This was then compared to solving the problem using math.

Question:
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.

Procedure:
By using Newton's second law, we were able to get acceleration as a function of time and the function looked like so:

By using calculus, we know that by taking the integral of a(t), we would get the function of v(t) which would look like so:

 Finally, in order to get x(t) we would need to take the integral of v(t). The resulting function would look like so:

Then, we can find time by finding the time at which v=0. We can then plug this calculated time into our x(t) equation to calculate how far the elephant will travel.

The work for this has been done below:

Numerically:
Later, we were shown in lab another way of solving the problem by using the excel program on our computers. We labeled the columns like so:

1) Column 1: t - time that increased by an increment of 0.1 for 258 rows
2) Column 2: instantaneous acceleration by using the formula of net force of mass for a certain time
3) Column 3: average acceleration given by the formula a_avg= (at + a(t+1))

                                                                                                                            2

4) Column 4: change in velocity given by (a_avg)*(delta t)

5) Column 5: velocity as a function of time given by the equation: v(t)= vn +(vn - 1)
6) Column 7: position as a function of time given by the equation:  
x(t)= (vn + vn - 1)*(tn - tn - 1 ) + xo                                                                                                            2
7) Column 6: change in position given by the equation x(n + 1) - xn

Next we search our data in column 5 (v as a function of time) until we find the point at which velocity goes from positive to negative.

Now that we have this value we change the interval in the time to both 1 s and 0.5 s and compare the data from both:
1 second:

 0.5 seconds:

As we can see, at 0.5 s, the distance traveled is 248.6971 m and at 1 second it s 248.3791 m. At the 0.5 second we are closer to our value calculated analytically of 258.7 m.

Conclusion:
At this point we would like to compare the values we got analytically vs numerically.

Analytically: x=248.7 m
Numerically: At t=.1s: 248.6927010m
                         At t= 1 s: 248.3791 m
                         At t=0.5 s: 248.6971 m

We chose our time intervals in excel to be very small until the difference between the two resulting values became negligible. In short, by doing this problem numerically, the benefits of having a program like excel and the ease with which one can do hard problems was demonstrated. Although both numerically and analytically used similar processes, I believe numerically was much easier, faster, and less tedious. Furthermore, if indeed done analytically, you could check your work numerically and vice versa.