09-March-2015: Non-Constant Acceleration Problem

The purpose of this lab is to illustrate the utility of the computer program "Excel" in complex physics problems that involve a non-constant acceleration.

The Problem:
For this lab, we were given a word problem involving a non-constant acceleration. The problem regarded a 5000-kg elephant on friction-less roller skates with a velocity of 25 m/s once it reaches the bottom of some hill. On the elephant is a rocket that produces a constant 8000 N thrust opposite the elephant's direction of motion. However, the problem continues, adding the fact that the mass of the rocket is a function of time m(t)= 1500kg - 20kg/s(t). Our goal is to find how far the elephant moves before stopping.

Fig. 1
Using integration leaves room for
far too much error.

Why The Problem is Difficult:
What makes this problem difficult is that all of our simple kinematic equations are now obsolete. Previously, we had always assumed acceleration to be constant. Thus, in order to find the distance traveled, we must use calculus to derive some function that relates position and time. We know that velocity is the derivative of position, and acceleration is the derivative of velocity. Thus, in order to find position, we have to integrate the acceleration function twice. However, this proves to be time consuming, tedious, and extremely vulnerable to multiple errors in calculation (See Fig. 1).


Numerical Integration:
To complete the problem, we are going to use the program "Excel" to manually integrate our values. In order to do this, we needed to open up a new spread sheet and enter the following at the top of the first 6 columns: time(t), acceleration(a), average acceleration(a_avg), change in velocity(Δv), velocity(v), average velocity (v_avg), change in position(Δx), and position(x). Above these, in cells A1 and B1, we indicated the time interval(Δt) to be 0.01 seconds (see Fig. 2)

Fig.2
These columns will allow us to recognize what each number means once we begin to input our equation and values.

The goal of making this spreadsheet was to find a relationship between all these different features of motion. If we provided "Excel" with the basic equations that described these things, then the program would be able to calculate thousands of them in seconds. All we have to do is input the simple things about motion we already know.

We Know:

• With this given problem, acceleration = -400/(325-t). 
• The change in velocity is the product of average acceleration and the time interval. 
• The velocity at any time "t" is just the initial velocity added with the change in velocity. 
• The change in position is the product of average velocity and the the time interval.
• The position at any time "t" is just the initial position added with the change in position.

To input this information, we have to provide each column with their own equations. For time(t), we simply start at zero and add the time interval(Δt) to each previous time. For acceleration(a), we input the formula -400/(325-t). The average acceleration(a_avg) will be calculated by the previous two accelerations. For the change in velocity(Δv), we multiply the average acceleration(a_avg) and the time interval(Δt). For the velocity(v) at that specific time, we simply add the change in velocity(Δv) to the previous velocity and in the case of this problem, we are given an initial velocity of 25 m/s. The average velocity(v_avg) is again calculated by the previous two velocities. To find the change in position(Δx), we multiply the average velocity(v_avg) by the time interval(Δt). Finally, the position(x) is calculated by adding the change in position(Δx) to the previous position and in the case of this problem, we start at a position of 0m. We input all these equations into excel, and from here we can now calculate the values we are looking for. 

For the problem, once the velocity was zero, we found the position to be 248.697m with a time interval(Δt) of 0.05s. (See Fig. 3)

Fig. 3
Once we had our time interval, we continued down the spread sheet all the way until velocity was
as close as it could be to zero. We then looked for the position at that particular time.

Conclusions:

1. When doing the problem analytically, we found the exact answer for the problem, which was 248.7m when rounded. Our value of 248.697m had a percent error of 0.001%. If we had chosen an even smaller interval, our answer would have been even more exact. 
2. The issue is now knowing when the time interval is small enough to yield you a usable answer. The only way to know this is to continuously input different values for Δt until the difference between the answers you are looking for become negligible.