We might be tempted to make the small extra time equal to zero. But in our
thought experiment, the stone has not moved at all. It has moved no distance
and taken no time to do it! This would give us the average speed 0/0 which the
Irish philosopher Bishop Berkeley famously described as the ‘ghosts of departed
quantities’. This expression cannot be determined – it is actually meaningless. By
taking this route we are led into a numerical quagmire.
To go further we need some symbols. The exact formula connecting the
distance fallen y and the time x taken to reach there was derived by Galileo:
y = 16 × x^2
The factor ‘16’ appears because feet and seconds are the chosen measurement
units. If we want to know, say, how far the stone has dropped in 3 seconds we
simply substitute x = 3 into the formula and calculate the answer y = 16 × 3^2 =
144 feet. But how can we calculate the speed of the stone at time x = 3?
Let’s take a further 0.5 of a second and see how far the stone has travelled
between 3 and 3.5 seconds. In 3.5 seconds the stone has travelled y = 16 × 3.5^2
= 196 feet, so between 3 and 3.5 seconds it has fallen 196 − 144 = 52 feet.
Since speed is distance divided by time, the average speed over this time interval
is 52/0.5 = 104 feet per second. This will be close to the instantaneous speed at
x = 3, but you may well say that 0.5 seconds is not a small enough measure.
Repeat the argument with a smaller time gap, say 0.05 seconds, and we see that
the distance fallen is 148.84 – 144 = 4.84 feet giving an average speed of
4.84/0.05 = 96.8 feet per second. This indeed will be closer to the instantaneous
speed of the stone at 3 seconds (when x = 3).
We must now take the bull by the horns and address the problem of
calculating the average speed of the stone between x seconds and slightly later at
x + h seconds. After a little symbol shuffling we find this is
16 × (2x) + 16 × h
As we make h smaller and smaller, like we did in going from 0.5 to 0.05, we
see that the first term is unaffected (because it does not involve h) and the
second term itself becomes smaller and smaller. We conclude that
v = 16 × (2x)
marcin
(Marcin)
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