Kevin Tran
Lab Partners: Kevin Nguyen, Jose Rodriguez
March 11, 2017
         

Non- Constant acceleration problem/Activity

          Purpose: Find out how far the elephant goes before coming to rest.

          Introduction: A 5000-kg elephant is going 25 m/s on frictionless roller skates. It will go down a hill and arrive at ground level. A 1500-kg that is attached to the elephant's back will generate a constant 8000 N thrust opposite to the elephants direction of motion. In addition the rocket is burning fuel at a rate of 20 kg/s. To find the distance that the elephant traveled before rest, we need to find the time at which the velocity of the elephant is zero.

Newton's 2nd law gives us the acceleration of the elephant + rocket as a function of time (Picture on the left).

In the picture below are our results of all the calculations for each variable.

We will now look into the data more carefully. We used the numerical approach to get the distance that the elephant traveled.

First, we used a set of numbers in the time column, but we decided to use the first 24 seconds to guess where the elephant might be within the 24 second range.

In the next column(picture below), we found our accelerations for each second interval within the 24 seconds. By applying the acceleration formula that we found (formula at top of the page), we are able to find the acceleration for each second. Since we know that the rocket is trying to push the elephant in the opposite direction, the acceleration will be negative.

Next, we have the chart for delta v and velocity within the 24 seconds time interval. 

In order to find delta v, we use the formula to find delta v. (picture below). To find delta v, we use the accelerations from one time to another. 

Here is an example to find delta v1 (picture below). I use the collected data from my acceleration chart and applied it into the equation.

To find velocity, we add up delta v and velocity. Since the elephant has an initial velocity of 25 m/s, the velocity at t=0 is still 25 m/s. As every second progress, we add the current velocity at each second interval with the delta v at each time interval.

For example:

Next, we have the average acceleration chart at each time interval of 24 seconds. (picture below). To find average acceleration, we use the formula provided below.

Next, we have the chart for delta x and position for the first 24 seconds. Since we know that the elephant is at the starting point, when t= 0, x= 0 m

To find delta x, we use the formula to find delta x. (picture below).

Here is an example to find delta x1 (picture below). I use the collected data from my velocity chart and applied it into the equation.

To find distance, we add up delta x and distance. Since the elephant is at the starting point, it is starting at x=0. As every second progress, we add the current distance at each second interval with the delta x at each time interval.

For example: 

Next, we have the average velocity chart at each time interval within the 24 seconds. To find average velocity, we use the formula provided below. 

To recap, in order to find the distance that the elephant traveled, we have the know when velocity=0. In between the 19th and 20th row on the velocity column, we can see the velocity is a little above 0 (0.90392744) and below 0(-0.405405). Now, we look into the x column. In between 19 and 20 seconds, we took both x values and divided by 2 to find an approximate distance and we have come to a conclusion that the elephant traveled approximately 248.50 meters.

Analysis:

1. In comparison to our numerical approach and the analytical approach, we have calculated and got a result of 248.50 meters by using the numerical approach. By analytical approach, we have got the result of 248.7 meters. In comparison to the two results, they were .20 off from each other but incredibly close to each other.

2. In order to know when the time interval we chose for doing the integration is "small enough" it can be found when time is closest to 0. If we did not have our analytical results, we would have to reduce the time interval from 1 second to something smaller like 0.01 seconds.

3.When the elephant initial mass is 5500kg, has a rocket mass of 1500 kg, has the fuel burn rate of 40 kg/s and the thrust force is 13000 N at 25 m/s, the elephant distance is 164.03 m by doing the analytical approach. After doing all the steps, I have plugged in my time of 12.96 seconds into the expression of x(t). (picture below).

          In conclusion, we were able to find out how far the elephant has traveled. Around 19 seconds, the elephant has traveled 248.50 meters according to our numerical approach. This lab has taught us how to find the distance of anything with a certain amount information. We were able to apply numerous formulas to find the results that we needed.