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

Chapter 3: Meanwhile back on the Cheddar ...
We have seen how the velocity v of the Cheddar could change with time, but how about the distance d through which it falls? This is a bit trickier since velocity is a continuously changing quantity (but isn't that also true of the increase of the bacteria?). However, we could reckon that the distance fallen in any second is given by its average velocity in that second (why?) and so build up a new table:

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?

----------------------------------------------------------------------------------------------------------------------

The additional differential equation is just [image], or s' = gt, with s = 0 when t = 0. The difference equation [image] 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 gt2 + c , and, from the initial condition,
c = 0 so s = 0.5 gt2 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 s' = gt .
Working with lists we have to use a little ingenuity!

[image] [image]
[image] [image]

[image] [image]
[image] [image]

[image]

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:
[image] [image]

[image]

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.