March 4, 2015: Non-Constant Acceleration

Non-Constant Acceleration

Purpose:

The purpose of this lab was to make us see physics and math in a different way. By this I mean that there are multiple ways to get to an answer. We found this out when we compared our integrated answer with that of our linearized excel answer and we found that when our linerization intervals are just small enough, the answers are similar.

Problem:

 We have an elephant and rocket with a combined mass of 1500kg rolling on a friction-less road at 25m/s. We are trying to find the distance it will take to slow down the elephant in the rocket if the rocket is providing thrust in the oposite way at 400N.

Missing here is 400N in the negative direction.

 Solving the Problem:

To solve the problem we decided to integrate the function of acceleration since it should give us the velocity function. Then we integrated the velocity function as it should give us the displacement.

1. We integrate a(t) and get v(t).

2. Once we get v(t), we solve to find the time at which velocity =0.

3. Then we integrate our v(t) to obtain our position formula.

4. Then we plug in the time into our position equation and find out the distance traveled by our rocket-boosted friend. 

Excel Way:

1. First we set up our 8 columns and labeled them "t" for time, "a" for acceleration, "a_avg" for average acceleration,

2. We found that the most accurate acceptable answer would be found using increments in time of 0.01 seconds, so thats what we set the time to be. Then we set our acceleration function equal to the one in the problem which was a(t) = -400/(325-t). The average acceleration was the acceleration at a time plus the acceleration of the time prior over two. 

3. Next we input the average acceleration multiplied by time to find the change in velocity.

4. After being given a change in velocity we used that multiplied by time to obtain and added that to our velocity to find our final velocity. 

5. To obtain our velocity we simply add our velocity to our change in our velocity.

6. We then add our velocity to the previous velocity and divide that by two to find our average velocity,

 7. To get our change in position we simply multiply our average velocity by our time gap 0.01

8. Finaly to obtain our distance we added our change in position

9. Now we followed the numbers until the moment before our velocity becomes negative to find the stopping point.

10. Our answer is the time frame at which velocity becomes negative, or in other words the closest it gets to 0. In this case the time we got was 19.69 seconds.

Summary:

At the end of the lab experiment it was evident that doing this problem by hand is longer and harder than having the computer (excel) do it. The gap was small enough such that our answer and that from excel were identical. It would be possible to get a closer answer but the difference at this point is negligible and good enough for our purposes.

Answers to lab questions

1.  Doing this problem analytically is do-able but not the shortest and easiest way. The benefits of doing it this way however is that we can achieve an exact answer. Doing this problem numerically however is easier than by integration.
2. When our velocity is as reasonably close to zero, in our case up to 2 decimal places.