Ian Lin March 8, 2017
This experiment is trying to analyze the numeric and analytical approach to solving when an object in question reaches zero velocity when the acceleration is not constant.
Most acceleration underwent by objects in the world are not constant. In fact, even objects accelerating under the influence of gravity isn't constant due to air resistance. Therefore, when determining the three crucial variables of acceleration, velocity, and position, one has to use calculus. However, some functions that express acceleration as a function of time is complex and hard to integrate. This gives rise to the numeric approach of using an excel spreadsheet to measure accurately measure acceleration, velocity, and position.
We entered the relevant information onto the excel spreadsheet and let the computer compute the values. The initial variables are the mass, velocity, burn rate, force, and change in time. Then the columns were filled down with time, acceleration, average acceleration, change in velocity, velocity, average velocity, displacement, and position. That's basically all there is to it.
| Mo | 6500 | kg | a given by | F / (Mo - bt) | |||
| Vo | 25 | m/s | (-8000) / (6500 - (1500-20)t) | ||||
| b | 20 | kg/s | |||||
| F | -8000 | N | |||||
| Δt | 1 | s | |||||
| t | a | a_avg | Δv | v | v_avg | Δx | x |
| 0 | -1.230769231 | 25 | 0 | ||||
| 1 | -1.234567901 | -1.232668566 | -1.232668566 | 23.76733143 | 24.38366572 | 24.38366572 | 24.38366572 |
| 2 | -1.238390093 | -1.236478997 | -1.236478997 | 22.53085244 | 23.14909194 | 23.14909194 | 47.53275765 |
| 3 | -1.242236025 | -1.240313059 | -1.240313059 | 21.29053938 | 21.91069591 | 21.91069591 | 69.44345356 |
| 4 | -1.246105919 | -1.244170972 | -1.244170972 | 20.04636841 | 20.66845389 | 20.66845389 | 90.11190745 |
| 5 | -1.25 | -1.24805296 | -1.24805296 | 18.79831545 | 19.42234193 | 19.42234193 | 109.5342494 |
| 6 | -1.253918495 | -1.251959248 | -1.251959248 | 17.5463562 | 18.17233582 | 18.17233582 | 127.7065852 |
| 7 | -1.257861635 | -1.255890065 | -1.255890065 | 16.29046613 | 16.91841117 | 16.91841117 | 144.6249964 |
| 8 | -1.261829653 | -1.259845644 | -1.259845644 | 15.03062049 | 15.66054331 | 15.66054331 | 160.2855397 |
| 9 | -1.265822785 | -1.263826219 | -1.263826219 | 13.76679427 | 14.39870738 | 14.39870738 | 174.6842471 |
| 10 | -1.26984127 | -1.267832027 | -1.267832027 | 12.49896224 | 13.13287826 | 13.13287826 | 187.8171253 |
| 11 | -1.27388535 | -1.27186331 | -1.27186331 | 11.22709893 | 11.86303059 | 11.86303059 | 199.6801559 |
| 12 | -1.277955272 | -1.275920311 | -1.275920311 | 9.951178622 | 10.58913878 | 10.58913878 | 210.2692947 |
| 13 | -1.282051282 | -1.280003277 | -1.280003277 | 8.671175346 | 9.311176984 | 9.311176984 | 219.5804717 |
| 14 | -1.286173633 | -1.284112458 | -1.284112458 | 7.387062888 | 8.029119117 | 8.029119117 | 227.6095908 |
| 15 | -1.290322581 | -1.288248107 | -1.288248107 | 6.098814781 | 6.742938834 | 6.742938834 | 234.3525296 |
| 16 | -1.294498382 | -1.292410481 | -1.292410481 | 4.8064043 | 5.45260954 | 5.45260954 | 239.8051392 |
| 17 | -1.298701299 | -1.29659984 | -1.29659984 | 3.509804459 | 4.158104379 | 4.158104379 | 243.9632435 |
| 18 | -1.302931596 | -1.300816447 | -1.300816447 | 2.208988012 | 2.859396236 | 2.859396236 | 246.8226398 |
| 19 | -1.307189542 | -1.305060569 | -1.305060569 | 0.903927443 | 1.556457727 | 1.556457727 | 248.3790975 |
| 20 | -1.31147541 | -1.309332476 | -1.309332476 | -0.405405034 | 0.249261204 | 0.249261204 | 248.6283587 |
| 21 | -1.315789474 | -1.313632442 | -1.313632442 | -1.719037475 | -1.062221254 | -1.062221254 | 247.5661375 |
Acceleration
acceleration= F( net) / m(t)= -8000 n /6500 kg- 20 kg/s * T
acceleration (1)= -8000 n /6500-20* 1 = -1.234567901 m/s^2
Change in Velocity
Δv= average acceleration *Δt
Δv= -1.234567901 m/s^2 * 1s=-1.234567901 m/s
Velocity
velocity= Δv+ v ( initial)
Velocity (1 s)= -1.23 m/s + 25 m/s= 23.77 m/s
Δ x
Δx =velocity average *ΔT
Δx ( 1 s ) = 24.38 m/s * 1 s = 24.38 m
This lab didn't involve any physical measurement of any processes at all. All of the lab was on paper and excel spreadsheet. The problem was to figure out how far the elephant that had a variable acceleration with time would go before slowing down to a stop. The analytical approach is to use integral calculus to figure out velocity and position and then setting velocity of the elephant equal to zero and then plugging that time into the position function to find the displacement. The excel spreadsheet method allowed the computer to calculate numerically to great precision I might add, the distance that the elephant travels before coming to rest. The two answers from the two methods agree closely to a tenth of a meter.
1. The two answers agree closely: analytical answer: 248.7 m , numerical answer: 248.9 m
2. The time interval for doing the integration is small enough when there're a bunch of velocity values that are close to 0 by 0.1 m increment, and the distance values that correspond to the velocity values differ by 0.1 m increment. When the distance values appear to differ by smaller and smaller values so that the precision is sufficiently small, then the numerical result is good enough.
3. The elephant would go about 129 meters given the conditions.
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