We try to keep our brains sharp around here. Those who know physics can see
that this problem, which I plan to present to my friends' son who is attending
college right now is just a little brain exercise, but I would also like to
know if I came up with the right answer. Part of the calculation bears
similarity to figuring out paralell resistance and series capacitance, so,
while a bit OT it is not out in left field.:
________________
An amusement park ride accelerates in a horizontal direction to a speed of 120
mph in 4 seconds. What is the average G effective force experienced by the
rider during that time ?

What is the force vector, using that figure, as compared to a line
perpendicular to the ground ?
_________

I came up with 2 2/3rd Gs. If necessary, I'll "show my work" in a followup.

The second part I have not yet calculated, but my guess is around 52 degrees.

To me, this is much better than discussing the last episode of "Friends" or
some dumb crap like that.

JURB

On 10 Aug 2003 20:25:01 GMT, (JURB6006) put finger to
keyboard and composed:

An amusement park ride accelerates in a horizontal direction to a speed of 120
mph in 4 seconds. What is the average G effective force experienced by the
rider during that time ?

v = u + a*t

where v = final velocity
u = initial velocity
a = acceleration
t = duration

Substitution gives:

120 * 1610 / 3600 = 0 + a * 4

Therefore a = 1610 / 30 /4 = 13.4 m/s2

Now 1G = 9.8 m/s2, so a = 13.4/9.8 = 1.37G

What is the force vector, using that figure, as compared to a line
perpendicular to the ground ?

|\
| \
1G | \
| \
V \
---------
1.37G

The force vector's magnitude is sqrt( 1 + 1.37 * 1.37) = 1.70G.

The angle relative to the vertical is arctan(1.37) = 54 deg

If you perform the same calculation in imperial units (yuck), then you
have:

120 * 5280 / 3600 = a * 4

So a = 5280 / 30 / 4 = 44 ft/sec2

Now 1G = 32 ft/sec2 , so a = 44/32 = 11/8 = 1.375G


- Franc Zabkar
--
Please remove one 's' from my address when replying by email.