08-March-2017: Non-Constant acceleration problem/ Activity

Title of Lab: Non-constant acceleration problem/ Activity
Kevin Nguyen
Lab Partners: Kevin Tran, Jose Rodriguez
Date of Lab performed: 08-March-2017

Purpose: The point of this lab is to learn how to solve a physics problem by using excel spread sheets.

Theory/ Introduction: In this lab, we were presented with a problem pictured below.

To solve for the distance the elephant travels before coming to rest, we could either take the analytical approach (which takes a massive amount of time) or we could take the numerical (arguably faster) approach. In the numerical case, we need several variables in order to create a spreadsheet. We used total mass of elephant plus the rocket on it, which was 6500 kg, the initial velocity of the elephant, which was 25 m/s, the burn rate of the rocket, which was 20 kg/s, the total force exerted on the elephant (in the opposite direction) initially, which was 8000N, and the change in time, which we can arbitrarily change, in order to solve for the position of the elephant using values of acceleration, average acceleration, change in velocity, average velocity and change in position. From Newton's second law, we used an acceleration function a(t) = net force / (total mass minus burn rate multiplied by time) in order to solve for the variables needed to find the position of the elephant.

Summary
At first, we set up the spreadsheet so it can look like the picture below.

Then we plugged in the values corresponding to their boxes. We then entered the function into the box below "a" and used the value given to solve for the other variables such as a_avg, ∆v, v, v_avg, ∆x, and x. After we finished setting up the values, we "filled" the information down. We changed the value of ∆t to 1 second, 0.1 second, and 0.05 second, showing the values below.

∆t = 1 second

∆t = 0.1 second

∆t = 0.05 second

To solve for the position of the elephant, we filled the information down and looked where velocity approximately equaled zero to find where did the elephant ended up.

The highlighted portion shows the position of the elephant when velocity equals zero.

After finding where velocity approximately equaled zero, we found that at t = 19.65 to 19.70 seconds, the position of the elephant was at approximately 248.70 meters.

Table of measured data:

Table of calculated results:

Calculated results for ∆t = 1 second

Explanation of analysis:

Addressed in the theory/ introduction section.

Conclusion:
Using the numerical approach actually made this physics problem much easier and faster than if we were to solve this problem using the analytical approach. We also got the same answer as the answer given in the lab manual.

Questions
1. We found our answer and compared the result we got to the answer in the lab. We confirmed that the answer we got matched the answer in the lab, confirming that the numerical approach also works in finding the answer to the problem.
2. We knew when the time interval was small enough when only the thousandths digit of the x value was changing each ∆t.
3.

Using the numerical approach, the answer that we got was 164.04 meters.