8-Mar-2017
Idea: Find how far a 5000 kg elephant traveling 25 m/s goes before coming to rest when a 1500 kg rocket mounted on its back generates a 8000N thrust opposite to the elephant's direction of motion.
Summary: In Excel, we first created a column where we could enter the given values, and be able to adjust them later on.
- Column A stood for my intervals of time, Column B for the acceleration of the rocket, column C for the average acceleration, Δv for Column D, V for Column E, Δx for Column G, and X for Column H. Because the acceleration isn't constant, I had to calculate Δv, v, a, Δx, and x separately, as shown below. In the first case, acceleration was calculate by dividing the net Force (8000N) by 6500-20t. Thus a= (400)/(325-t)
- Then, I dragged the columns down until my value of V in Column E was zero.
Analysis
When I stretch the columns, for each of the intervals of time I set (1s, .1s, and .05s) I look for values of t where V is zero. As I saw when ΔT was 1, V is zero in between T = 19 and 20. As I decreased ΔT from 1 to .1, to .05, the value of V was narrowed down to be most likely to between T=19.65 and T=19.7s.
Data (Time Interval 1)
Data (Time Interval 0.1)
Data (Time Interval 0.05)
Conclusions/Questions
1) Analytically, the value for of T where ΔV was 0 is 19.69075s. My numerical result was a value between 19.65 and 19.7s. My numerical value was rather accurate.
2) I know if my interval is "small enough: once the values of V are similar to a few decimal places. If I didn't have the analytical approach, I would have to keep decreasing my ΔT values and make a judgment as to when I would accept the margin of error.
3) If the elephant's initial mass were 5500 kg, the fuel burn rate is 40kg/s and the thrust force is 13000N, the new acceleration equation is:
Plugging this formula back into Excel along with our new data,
The new V=0 in an interval of .05s gives me a value of time between 12.95-13s.
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