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# i actually jumped backwards when i saw that!
very woo!

PS:
im needing help with my physics homework.....how do you work out acceleration on a speed/time graph? (with a straight line)
(Lord of the Beans, Sat 1 Nov 2003, 13:42, archived)
# acceleration
on a space hopper?
(I helped save b3ta! Sunshine Elephant, Sat 1 Nov 2003, 13:43, archived)
# it could be on a space hopper yes
.
(Lord of the Beans, Sat 1 Nov 2003, 13:44, archived)
#
i love spacehoppers

(I helped save b3ta! Sunshine Elephant, Sat 1 Nov 2003, 13:45, archived)
# yay! me too...
(I helped save b3ta! eclectech is older but no wiser, Sat 1 Nov 2003, 13:52, archived)
# He doesn't
look very happy to be there, poor little mite.
(I helped save b3ta! splorp ®, Sat 1 Nov 2003, 13:53, archived)
# Yay me too!
I've heard there's a space hopper in these pics somewhere...

(I helped save b3ta! Evil Prof. Pixel Masher I made this!, Sat 1 Nov 2003, 15:44, archived)
# hahahahahaaa
hahahashahaha my sides hurt !! hahahah
(BULLSHIT_and_ONIONS, Sat 1 Nov 2003, 18:18, archived)
# hmmm,
gonna dream nice tonight
(bwana dik, Sat 1 Nov 2003, 22:55, archived)
# isn't it
speed divided by time?
or something?
(I helped save b3ta! Lora, Sat 1 Nov 2003, 13:44, archived)
# ask and ye shall recieve
its the gradient of the graph.....gah, i can't believe i remember that
(Vin diesel touches my tra la la shove a bastard in it, Sat 1 Nov 2003, 13:44, archived)
# er..
speed x time = distance.

here sums upthe entirety of my physics knowledge.
(I helped save b3ta! Prof. Slocombe, Sat 1 Nov 2003, 13:46, archived)
# The solution
Is the derivative, young grasshopper. Or the average slope between two points, for you non-calculus folk.
(+3 Baroque Vambrace of Icthyology, Sat 1 Nov 2003, 13:46, archived)
# so
it would be the lowest add the highest divided by the time?
(Lord of the Beans, Sat 1 Nov 2003, 13:48, archived)
# right
it starts at 0m/s and goes to 15m/s in 20 with a steady acceleration. how do i work it out?
(Lord of the Beans, Sat 1 Nov 2003, 13:50, archived)
# change in speed
divided by time.

15/20 = acceleration
(I helped save b3ta! Sammo, Sat 1 Nov 2003, 13:52, archived)
# to help you
it's 0.75 m/s/s
(I helped save b3ta! Sammo, Sat 1 Nov 2003, 13:52, archived)
# thanks
.
(Lord of the Beans, Sat 1 Nov 2003, 13:56, archived)
# google is your friend.
(I helped save b3ta! dipdapdop, Sat 1 Nov 2003, 13:52, archived)
# blimey, that took a long time to appear!
(I helped save b3ta! Kamikaze Stoat £4.09, Sat 1 Nov 2003, 13:53, archived)
# err, long time ago but
v=u+at
so a=(v-u)/t
final velocity- start velocity over time
(I helped save b3ta! e.parsley, Sat 1 Nov 2003, 13:53, archived)
# er...
If it's a straight line, all you need is the slope of that line. Rise over run. (y2-y1)/(x2-x1)=m=acceleration
(+3 Baroque Vambrace of Icthyology, Sat 1 Nov 2003, 13:53, archived)
# wouldn't you have to
work out the acceleration for each time segment and find the average of all those results?
(I helped save b3ta! Kamikaze Stoat £4.09, Sat 1 Nov 2003, 13:51, archived)
# ooohhhh
i cant be arsed. ill get my brother to do it for me. i think he has a physics fetish or something
(Lord of the Beans, Sat 1 Nov 2003, 13:52, archived)
# was it you who was looking for the glasscock pic?
well it's here anyhow...
www.businessladiesgolf.com/Photos/trophy_kiss.gif
(I helped save b3ta! eclectech is older but no wiser, Sat 1 Nov 2003, 13:55, archived)
# lol
me likes
(Lord of the Beans, Sat 1 Nov 2003, 13:57, archived)