4 March 2015. Non-Constant Acceleration (how to integrate easier)

Purpose
to find an easier and faster way using numerical integration (excel) instead of doing an analytic integration in a problem with a non-constant acceleration symbolized as a function of t (time).


The Question:

A 5000 kg elephant on frictionless roller skates with V=25m/s is going to the bottom of a hill. When it reaches the bottom, a rocket in his back generates F=8000 N opposite direction of the motion.
The mass of rocket changes with time, because the fuel inside it burns at a rate of 20 kg/s. Therefore, we have: m(t) = 1500 kg - 20 kg/s x (t).
Find the distance until the elephant reach Vt=0m/s.

Using analytic Integration:

• shown in the picture below.

• Now, we just insert the number. For the elephant to come at rest, the final velocity = 0 m/s.

• We have t = 19.69075 s and X = 248.7 m.

Using numerical Integration:

• First, we have to understand the concept written in the picture below.

• Next, we compute this to excel and vary the ∆t.
• With time interval of 1 s, we got:

• With time interval of 0.1 s, we got:

• With time interval of 0.05 s, we got:

• With time interval of 0.02 s, we got:

Conclusions:

• Using numerical integration (excel) is much easier because the program itself do not care if we want to smaller the ∆t into per milliseconds. It is going to generate the result in just a fraction of a second. Meanwhile, using analytic integration is really risky. At first, we do not even know whether the equation itself is able to be integrated or not. Moreover, it takes time and generates many opportunities to make mistakes along the way.
• But, when we need to stop "smaller-ing" the ∆t? We know when to stop if the smaller value of ∆t does not generate much of a difference in the result (in this case x, the distance).
• Take a look when ∆t = 1 s. It generates t = 20 s and x = 248.628 m, when v=0 (elephant at rest).
• Then when we change ∆t to 0.1 s, we got  t = 19.7 s and x = 248.68 m, when v=0.
• Then we change the ∆t it again to 0.05 s, we got t = 19.7 s and x = 248.69 m, when v=0. At this point, the value of the result does not change much. This is the right time to stop.
• But, I did change the value of ∆t again to 0.02 s and got the exact same answer to six significant figures for the distance. This is to show that it is a little bit pointless to do the procedure one more time.
• The result that we get from doing both numerical and analytic way is more or less the same, to an extent when ∆t is minimize, with the differences (∆t=0.02 s) for the distance = 0.002 m.