6-March-2017: Non-Constant Acceleration Problem(Elephant)

Lab#3: Non-Constant Acceleration Problem
May Soe Moe
Lab Partners: Ben Chen, Steven Castro
Date Performed: 6-March-2017

Objective: To find out the distance of the elephant before coming to rest by numerically and analytically and compare the results

Problem with Given Information

Theory/Introduction: We were given the problem, where we wanted to find out how far the elephant goes before coming to rest, meaning we want to find the distance x when the final velocity v is zero. We knew the following information: the mass of the elephant on frictionless roller skates (5000-kg), the velocity of the elephant (25 m/s), the mass of the rocket on the elephant's back (1500-kg), the rocket's thrust (8000N), and the function m(t)= 1500kg-(20kg/s)*t. In order to figure out the distance, we will approach this problem analytically and numerically. For analytical approach, the professor already did a bunch of calculus, integrating, to find the velocity and position function from Newton's Second Law, Fnet=m(t)a. To numerically approach, we used Microsoft Excel to put in the known values and find what we do not know to figure out the distance that we are looking for. We tried to figure it out by letting the Excel calculate the data and see when the velocity is zero from the spreadsheet. Then we compared what we got analytically and numerically. 

Summary:

(1) We put in the known values from the given problem into Microsoft Excel: M0=6500 kg, v0= 25m/s, b(burning fuel rate)=20kg/s, Fnet= -8000N. We will set Delta t to be 1 second at first and calculate the acceleration, change in velocity, velocity, change in position, and position at each time interval.

(2) What we wanted to find out was the distance the elephant traveled when its final velocity is zero. Since we don't know when the elephant is at rest, we tried to figure out the time when the velocity is zero by filling down the row. 

(3) The concept behind is that we wanted to find out the time it took for the elephant to come to rest because we were using trapezoidal rule to approximate the distance. By making the time interval smaller, the trapezoid gets smaller and it makes accurate approximation of the distance.

How we got the equation for Change in Velocity. 

How We Got the equation for Change in Position.

(4) After filling down the rows, we tried to see when the velocity was zero during the 1 second time interval.

(5) When the time interval is 1second, we saw that the velocity changed from 0.90392744m/s at t=19s to -0.405405m/s at t=20s. This means that the velocity reached zero between t=19s and t=20s. 

(6) As we wanted to get better understanding of exactly when the velocity is zero, we changed the time interval to 0.1s.

(7) We saw that velocity changed from 0.11888376m/s at t=19.6s to -0.0121135m/s at t=19.7s, which gave us more accurate time period and the change in velocity than when time interval was 1 second.

(8) We tried to make Delta t smaller by putting 0.05 second instead of 0.1 second to see how much difference it made.

(9) From the screenshot, we could see that the velocity must be zero in between t=19.65s and t=19.7s since the velocity went from +0.05339083m/s to -0.0121131m/s. 

Conclusion:

(1) When we compared the results we got from doing the problem analytically and doing it numerically, we could see that it was pretty much the same. Analytically approach got us the distance x to be 248.7m. When we did the problem numerically, we plugged in time intervals to be 1second, 0.1 second, and 0.05 second. We got the distance x of 248.628m when delta t was 1 second, 248.698m when delta t was 0.1 second and 0.05s. 

(2) How we know when the time interval we chose for doing the integration is "small enough" is when delta t stops changing the distance x's value. When delta t is small enough, the distance's value will stay pretty much the same to several decimal places.If we didn't have the analytical result to which we could compare our numerical result, we could tell the distance by calculating standard deviation and see in what range we fall into, and to what percent we can be confident that we got the right answer.

(3) We did the same steps as above to figure out how far the elephant would go if its initial mass were 5500kg, the rocket mass is still 1500 kg, but now the fuel burn rate is 40kg/s and the thrust force is 13000N. We changed its delta t to be at 1 second, 0.1 second, and 0.05 second and got the distance x to be 164.021m, 164.034m, 164.034m respectively.