h2g2 The Hitchhiker's Guide to the Galaxy: Earth Edition

Entry: How To Calculate Your Reaction Times With A Ruler - A32877840
Author: SWL™ Never count your chickens until they've crossed the road. - U1775547

Just a quickie.

Nice idea, but I think you got your acceleration and terminal velocities confused - the ruler will start at 0 m/s, and accelerate at 9.8 m/s/s.

It will still be accelerating when you catch it.

So the math will be a bit more complicated. I don't have time now though,
back later.

vp

Yeah, I did wonder how quick the acceleration would be, but I figured it might not introduce too large a factor of error in such a rough & ready experiment.

How did I just know this one would be from SWL?

For an object initially at rest, the equation of motion is:

distance = 0.5 x acceleration x time x time, so

14 cms = 0.5 x 980 cms per sec per sec x time x time, or

time x time = 14 divided by 490 (0.5 x 980), so

time = square root (14 divided by 490) = 0.169 secs

or 169 milliseconds

Still pretty good; the average prior to middle-age is about 200 mS

I think the proper statement is: they fall at a constant acceleration rate of 9.8 meters per squared second - that is, 9.8 m/s per second, or: in a second the speed goes up 9.8 m/s.

If I remember correctly, the end formula if you drop something without initial speed is: h(t) = 1/2 g t^2, where h(t) is the height it has dropped during a time t, and g = 9.8 m/s^2.

Therefore, if h = 14cm = 0.14m you get t^2 = 0.28 / 9.8 , hence t = about 0.17 second.

Ok - as ever, I'll see if a concensus develops

It's calculus. Acceleration is the second derivative of distance with respect to time; you're trying to derive the time given the distance and acceleration. It's been far too long since university for me to remember how do do it, but your constant rate of fall is definitely wrong.

'nother nice little entry though, once the maths are sorted.

It's perfectly simple. You have
x = distance,
x' = speed = derivative of x w.r.t. time
x'' = acceleration = derivative of x' w.r.t. time.

'Cos Newton said so, we have x''= g (where g = 9.8m/s^2).

Therefore we obtain:
x'(t)-x'(0) = g(t-0) = gt,
and x'(0) = initial speed = 0 (otherwise your friend is cheating). Hence: x'(t)=gt.

Integrating once more we obtain:
x(t)-x(0) = 1/2*g*t^2 - 1/2*g*0^2,
and x(0)=0 because SWL said so.

Hence, x(t)=4.9*t^2.

Perfectly simple?

<>

Toy Box and I are saying exactly the same thing. He has expressed it in mathematical notation, whereas I tried to avoid it.

Plug in the numbers - 0.14 m or 14 cms, and 9.8 m/sec/sec or 980 cms/sec/sec - but stick to the same units metres or centimetres. You get the same result - 0.169 secs or 169 milliseconds.

If you change your:

<>

to:

<>

...you're home and dry

>>Perfectly simple?<<

I never mean it when I use this expression (which, by the way, always reminds me of Basil Fawlty).

I'm going with h5ringer's text

Bother - that formula line should have been:

The appropriate formula is: reaction time = square root(2 x distance fallen / 9.8)

and <>

should have been:

we divide 28 by 980 and take the square root.

Morning folks .

We still have a problem in third para of The Toy Box-h5ringer Solution:

<<>>


The ruler is not travelling at 9.8 metres per second, as it has not fallen for one second, but has in fact fallen for considerably less time. The math is good, but the phrasing is wrong.

My suggestion:

'We know that the ruler fell 14cm before it was caught. So, if the ruler was falling at 9.8 metres per second per second, or 980 cm per second per second, we divide 28 by 980 and take the square root. This gives a result of 0.169. So the reaction time was 0.169 seconds, or 169 thousandths of a second...'.


Is that better?
vp

For more accurate results, buy a longer ruler, and visit a planet heavier than this one. The acceleration due to gravity on Jupiter (woe betide anyone trying to stand on the surface of this puppy) is apparently 23.12 m/s/s accorinding to A402003 (although http://www.aerospaceweb.org/question/astronomy/q0227.shtml has it as 25.95 ).


vp

vp is quite right. That third para is being a right b**g*r.

I finally suggest:

<>
That's it, unless..

Getting there


<<>>

feels [less clumsy / more familiar] than gravity acceleration constant.

What do you think ?

I could live with it

I should hope so, it's genius