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
--
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