Purpose: Analytically and numerically solve a non-constant acceleration problem. Be able to successfully apply various formulas into an excel spreadsheet, as well as see the logic behind it, that would generate the same answer as when when solving the problem analytically.
Problem: A 5000 kg elephant on frictionless roller skates is going 25 m/s when its gets to the bottom of a hill and arrives on level ground. At that point a rocket mounted on the elephant's back generates a constant 8000 N thrust opposite the elephant's direction of motion. The mass of the rocket changes with time (due to burning the fuel at a rate of 20 kg/s) so that the
m(t) = 1500 kg - 20 kg/s*t
Find how far the elephant goes before coming to rest.
A simple drawing of the scenario. As shown in the picture, we are looking for x.
Analytically:
We are given the elephants acceleration as a function of time. Using Newtons second law, F=ma, we can rearrange to find a(t). The net force is (-) because it is in the direction opposite to the elephants motion. The total mass is the mass of the elephant + mass of the rocket.
We can next integrate a(t) from 0 to t to find ∆v to derive an equation for v(t).
After integrating a(t) to get v(t), we get the equation below.
Next we integrate v(t) from 0 to t to find ∆x to derive an equation for x(t)
Equation for x(t) below
Before finding how far the elephant goes before coming to rest, we need to find the time when the final speed of the elephant is zero. Solving for t from the equation below leads us to 19.6907 seconds.
Lastly, we plug t = 19.6907 s into our x(t) expression to yield x=248.7 m. This will be the distance the elephant travels before coming to rest.
Numerically:
Solving this problem analytically can be challenging, luckily we can do it with much more ease using excel.
We need to create columns as shown below.
1. Column A: Time begins at 0, so start by setting time increments by 0.1 seconds (= A3+$B$1)
2. Column B: Since acceleration is not constant, we need to input into B3 (= -400/(325-A3))
3. Column C: a_avg = (a1+a2)/2 we need to enter in C4 (=(B3+B4)/2)
4. Column D: ∆v = a_avg*t we entered in D4 (= C4*$B$1)
5. Column E: v = vf + vo. In E4 input (=E3+D4)
6. Column F: v_avg = (vf + vo)/2 in F4 enter (=(E3+E4)/2)
7. Column G: ∆x = v_avg*t in G4 we entered (=F4*$B$1)
8. Column H: xf = xo + ∆x lastly in H4 to find the final position we entered (=H3+G4)
We can now make our time increments smaller or larger to see if it make any difference in our answer. Below we see that if we used ∆ t = 0.1 seconds, at 19.6 seconds, our final position if 248.69m, which is pretty much exactly as we calculated analytically.
By making ∆ t = 1 second we still get a very close position as we calculated analytically, but its not exact.
Conclusions:
We are able to approach this by imagining a plot of a vs. t. If ∆t is small enough, then the curve between t0 and tf is approaching a line and ∆v between t0 and tf is the area under the curve of the a vs. t graph between t0 and t1.
We now imagine a graph of v vs. t. If ∆t is small enough we can approximate the curve between v0 and v1, v1 and v2 etc. as straight lines.
1.Doing the problem analytically and numerically yielded the same results. Numerically, the accuracy of the answer depended on the value we used for ∆t. If we used 0.1 as ∆t, we get matching numerical and analytical results. If we use 1 second, there is a small difference in the final position. In general the smaller we make ∆t, the more our analytical and numerical results would match.
2.You know the time interval analytically is small enough when the velocity becomes extremely close to 0. Another aspect to look at is when the difference in the distance values become close to 0 as well.
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