*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 _{} (a constant), or *v*' = *g,* with *v* = 0 when *t* = 0. The difference equation_{ } is just that of an arithmetic progression (additive sequence), where: .

_{ } . The solution of the difference equation is: _{} .

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:_{ } 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:

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 *

An alternative (usually preferable) approach is to use the statistical lists of the calculator or TII!

Now we can plot a scattergram of *vl* against *tl* and fit the function *v* = *gt* .

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:

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 .