Question

derived the third equation of motion v2=u2+ 2as

Answer 1

area under vt graph gives displacement
s= area under vt graph
s= ar(OABC)
s =1/2(BC+OA)*OC
s=1/2*(v+u)t
from 1st eq of motion t=v-u/a
s =1/2(v+u)(v-u/a)
s =v²-u²/2a
2as=v²-u²
u²+2as=v²
mark it as brainliest if helped

Answer 2

Hi mate...
Please see the attached files..
There's derivation of third equation of motion v2=u2+ 2as(graphically)




Here the derivation...( Not graphically)

(3) Third equation of Motion
v^2 = u^2 +2as

sol.
We know that
V = u + at
=> v-u = at
or t = (v-u)/a ………..eq.(3)

Also we know that
Distance = average velocity X Time
.: s = [(v+u)/2] X [(v-u)/a]
=> s = (v^2 – u^2)/2a

=>2as = v^2 – u^2


I hope my answer will satisfy you
#yahyaahmad#
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