Drop Tower 2
Author: Andreas Bauer

t_2

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t_1

You have calculated the average speed in the interval
from 3s to 4s. You also know that the gondola
reaches its top speed towards the end of its path.
Combining this, will the average speed from 3.5s to
to 4s be higher or lower than the average speed from
3s to 4s?
Experiment with interval length varying the blue point t_1
on the x axis and observe the change of the average
speed on the right. Is there a maximum to which
the average speed will rise or fall? Give reasons that
support your conjecture!

Because the gondola is steadily accelerating, its
average speed from 3.5s to 4s is higher than from
3s to 3.5s. When calculating the average of the entire
second, from 3s to 4s, the slower part from 3s to 3.5s
induces a lower average speed than what you would observe 
if you only considered the average speed from 3.5s to 4s.

Therefore, reducing the time interval length will increase
the gondolas average speed in the respective interval.

There is a maximum for this trend, however. Since the
gondola has a certain, limited speed at the end of the 
   ride, the average speed cannot increase indefinitely. 
      Instead, it approaches the momentary speed of 
         the gondola at the end of the ride, right before 
            the brakes set in.

A = Point(xAxis)
I = Intersect(xAxis, yAxis)
f = If(x ≥ 0, If[x ≤ 4, -(70 / 16) * (x + 4) * (x - 4)])
G = Intersect(f, xAxis, (4, 0))
j = Segment(I, G)
B = Point(xAxis)
C = (-1, 75)
D = (-1.2, 75)
Vieleck1, a_1, b, c, d = Polygon(A, B, C, D)
t = 0 (slider from 0 to 4, step 0.01)
E: (t, f(t))
a = false
s: a ∧ (t < 4)
hint = false
solution = false
F = Point(j)
h = OrthogonalLine(F, xAxis)
H = Intersect(f, h, (1.3976464739116345, 61.45380646108539))
i = Segment(H, F)
k: f(x(F))
Text1: "Average speed from t_1=" + (x(F)) + " to t_2=" + (x(G)) + ":" + "

$v_{avg} = \frac{h(t_1) - h(t_2)}{t_1 - t_2} = \frac{" + (f(x(F))) + " - " + (f(x(G))) + "}{" + (x(F)) + " - " + (x(G)) + "} = " + ((f(x(F)) - f(x(G))) / (x(F) - x(G))) + "$"
xF: x(F) - 4
Text2: "Average speed from t_1=" + (LaTeX[x(F)]) + " to t_2=" + (LaTeX[x(G)]) + ":

$v_{avg} = \frac{h(t_1) - h(t_2)}{t_1 - t_2} = \frac{" + (LaTeX[k]) + " - " + (LaTeX[f(x(G))]) + "}{" + (LaTeX[x(F)]) + " - " + (LaTeX[x(G)]) + "} = " + (LaTeX[k / (x(F) - x(G))]) + "$"
Vieleck2, e, g, l, m = Polygon((-1.46, f(t)), (-0.74, f(t)), (-0.74, f(t) + 3), (-1.46, f(t) + 3))