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# 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)