Jinx wrote:
> That's what I thought. If my Applied Maths (aaah Ms Pfannkuck,
> where are you now ?) memory serves me
>
> v^2 = u^2 + 2as
>
> v^2 = 0 + (2 x 9.81 x 3)
>
> v = 8m/s (29kph, 18mph)

It's not clear what exactly you're doing here since you didn't define your
terms, so let's start from the beginning.

  V = AT

Where V is the change in velocity as accelleration A is applied for time T.
The distance traveled during that time is the average velocity times the
time.  In our case the initial velocity is 0, and it's obvious from the
equation above that V changes linearly.  The average velocity of the
interval is therefore half the final velocity, which allows us to find the
distance D:

      1
  D = - VT
      2

We know the distance and accelleration and want to know the velocity, but
don't know nor care about time.  Solving for T:

      2 D
  T = ---
       V

Which can now be substituted in the first equation to find V:

      2AD
  V = ---      V**2 = 2AD     V = sqrt(2AD)
       V

  V = sqrt(2 * 9.8m/s**2 * 3.05m) = 7.7m/s

And this is assuming no air resistance.  1000G is 9800m/s**2.  The time it
takes to decellerate at 1000G is:

            7.7m/s
  Tstop = ---------- = 786uS
          9800m/s**2

So their claim implicitly says the card comes to a stop worst case in 3/4 of
a millisecond.  I guess that sounds plausible for a hard floor and
considering the hard plastic case.


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