Non-Constant acceleration problem/Activity Solution:

Objective:
To solve non-constant acceleration problem numerically and compare the result with the one solved by calculus.
Questions:
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 8000N thrust opposite the elephant’s direction of motion. 
The mass of the rocker changes with time (due to burring the fuel at a rate of 20kg/s) so that the m(t) =1500 kg - 20kg/st
Find how far the elephant goes before coming to rest.
Solving problem using calculus
First, we identified whether this question was related to constant acceleration or non-constant acceleration. Because the mass of the system is not constant, plus the system is affected by unequal force applied horizontally, this question is related to the concept of non-constant acceleration. With given data, we could find an equation for acceleration using the formula F=m*a which gave us:

a = F/m =           (-8000N)           

                                                     (5000 kg + (1500 kg + 20 Kg/s (t)) 

We had an acceleration equation and were asked to find the distance. Therefore, we took an integration to solve for v and then from v, we integrated again to find the distance. 
From an acceleration equation, we integrated to find a velocity equation. We knew that the initial velocity is 25m/s, we found a velocity equation by doing:

v = v_o + ⎰a(t)dt

After finding the velocity equation, we integrated one more time to find a position equation which helped us attain the final result. 

x = xo + ⎰v(t)dt

The detail of calculation is shown in the figure 1 below.

Figure 1: Using calculus to solve the question. First analyzing the question using the picture, then show steps how to integrate the function before attaining the final result.

         The final result we got for the distance x = 248.7m corresponding to the time t=19.7s. This means that the elephant will go 248.7m in 19.7s until it is coming to rest. 

Solving problem numerically.

            Instead solving this question by calculus, we also figured out another way to do it by using our Excel program. We set up a program which would give us a similar result to the one above. To start with Excel program, we first labeled column. From left to right, the order is time(t), acceleration(a), average of acceleration(a_avg), the change in velocity(△v), instantaneous velocity(v), the change of position(△x), lastly total distance(x). When we were done, we had a table like this:

Figure 2: The table with the order of time, acceleration, average of acceleration, change in velocity, instantaneous velocity, average of velocity, change in position, and instantaneous position.

      We then started to put data into the table. First we filled down the column of time up to at least 220 rows. Then we calculated the acceleration using an equation below

a = -400 / (325 - t² )

       Then to calculate the average of acceleration, we used an equation:

a_avg = (a_n + a_(n+1)) / 2

         Next, we calculated the change in velocity:

△v = (a_avg) * △t

       The results in change in velocity helped us find the instantaneous velocity:

v = v₀ + △v

       We then found the average of velocity:

v_avg = (v_n + v_(n+1)) / 2 

       To find the change in position afterward, we used an equation:

△x = (v_avg) * △t

       Finally, we came up with an equation to give us the final result which is the instantaneous position:

x = x_o + △x

       After entering the data and performing all the calculation, we got the table like Figure 3.

Figure 3: The table with data and result after performing all calculations.

      However, we would like to find the instantaneous position corresponding to the time t=19.7s, we continued to fill down the whole table until we reached the time t=19.7s. There were 200 rows we went through until we obtained t=19.7s. As you see in the Figure 4, the result we got for the position corresponding to t=19.7s was 248.92863m. However, we knew that there was some uncertainty in the  experiment, therefore we considered the range for the position from 248.90023m to 248.94392m.

248.90023m < x < 248.94392m

corresponding to the time: 19.6s < t < 19.8s

Figure 4: The results we got for the instantaneous position in the range of time from 19.6s to 19.8s
.

          We also noticed that every equation in our calculation depends on the value of time(t), thus we changed ∆t to see the effect. We first adjust ∆t= 0.05s, later ∆t=0.01s.

          At ∆t=0.05s, here is our results:

 Figure 5: The instantaneous position we got when adjusting ∆t=0.05s was 248.8439m

          At ∆t=0.01s, here is our results:

Figure 6: The instantaneous position we got when adjusting ∆t=0.01s was 248.72432m

         As we adjusted the value of ∆t, we noticed that the numerical result got closer to the analytical result which means that the effect of ∆t on the change of the instantaneous position was clear. The smaller the ∆t was, the more accurate the result was. 

Conclusion:

         In this lab, we learned two methods to solve a problem related to non-constant acceleration. One was using calculus with the integration technique, and the other was using spreadsheet. The results we got from doing the problem analytically and doing it numerically were pretty similar. They both gave us the instantaneous position equal 248m. The difference was the decimal places which could be explained by the rule of rounding number or the uncertainty we learned in the free fall lab. The time interval considered "small enough" was based on the difference between the analytical and numerical result. If the difference is really small or even negligible, we could state that the time interval we chose was "small enough." Consequently, we were able to get close results using different techniques.