b3ta.com board

Home » Messageboard » XXX » Message 2249204 (Thread)

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)

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

(

Sammo, Sat 1 Nov 2003, 13:52, archived)

to help you

it's 0.75 m/s/s

(

Sammo, Sat 1 Nov 2003, 13:52, archived)

thanks

(Lord of the Beans, Sat 1 Nov 2003, 13:56, archived)

google is your friend.

(

dipdapdop, Sat 1 Nov 2003, 13:52, archived)

blimey, that took a long time to appear!

(

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

(

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?

(

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?

(

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)