A Tale of two cheeses - Adrian Oldknow a_oldknow@compuserve.com

Chapter 1: The flight of the Cheddar - (expurgated version)

The acceleration of the Cheddar is that of gravity - i.e. g = 9.8 m/s/s approximately (as observed by Galileo one sunny afternoon in Pisa). Assuming that the balloon was stationary and that the cheese left the balloon at time t = 0 with downward velocity v = 0 then make up a table of its velocity v m/s after times t =1, 2, 3, .. . , s.

t 0 1 2 3 4 5

v 0 9.8
What would a graph of v against t look like? If the speed of sound in air (Mach 1) is about 330 m/s how long will it be before the Cheddar breaks the sound barrier? Is this likely? Give some reasons.
The differential equation is just [image] (a constant), or v' = g, with v = 0 when t = 0. The difference equation [image] is just that of an arithmetic progression (additive sequence), where: .
[image] . The solution of the difference equation is: [image] .

The anti-derivative of the differential equation is v = gt + c , and. from the initial condition, c = 0

so v = gt is the solution. If we integrate the differential equation we have: [image] with the same solution, and this is the area under the graph of v' = g .

We can model the number sequence using the SEQ plotting Mode of a TI graphic calculator:

[image] [image] [image]

An alternative is to use a spreadsheet - either Excel, or the spreadsheet of TI InterActive! (TII!) Here we can compare the scatterplot of the v sequence against t, and the graph of v = gt [image][image]
An alternative (usually preferable) approach is to use the statistical lists of the calculator or TII!
[image] [image] [image]
[image]
Now we can plot a scattergram of vl against tl and fit the function v = gt .
[image]
The area under graph v = g from t = 0 to t gives the change in velocity i,e, v = gt.

The time to reach the sound barrier would be:
[image] sec.
This seems very long - but we need to have some idea of the maximum height of an ordinary balloon - perhaps 1000m?? See
Adventure balloons .