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

*Chapter 3: Meanwhile back on the Cheddar ...
*We have seen how the velocity

*t* 0 1 2 3 4 5

*v* 0 9.8 19.6

*av* 4.9 14.7

*d * 0* *4.9 19.6

So how far (in metres) would the Cheddar have fallen before it reached Mach 1 ? What does a graph of *d *against t look like? What shape is it? Can you find formulae for the velocity *v* and the distance *d *after time *t*?

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*The additional differential_{ }equation is just _{}, or *s*' = *gt,* with *s* = 0 when *t* = 0. The difference equation_{ } is just that of a linear function, so its solution will be a quadratic.

The anti-derivative of the differential equation is *s* = 0.5 *gt*^{2} + *c* , and, from the initial condition,

*c* = 0 so *s* = 0.5 *gt*^{2 }is the solution. If we integrate the differential equation we have:_{ } with the same solution, and this is the area under the graph of *s'* = *gt* .

Working with lists we have to use a little ingenuity!

Now we can graph scatterplots of the data from *t, v* and *d*, and fit the quadratic model to *d*:

If the cheese could reach the speed of sound it would be after a time t secs given by:

The reason that the average velocity on the numerical data agreed exactly with the analytic solution is that we are finding the area under a linear function, so each strip is a trapezium, and the trapezium rule for integration is thus exact.