Question

derive 3rd equation of motion by graphical method....

please answer this question...

Answer 1

See this pic :)

If distance covered by body in time t and be s

S=Area enclosed b/w graph and time axis.

S=ar(OACD)

S=1/2 (AC+OD)×OA

S=1/2(V+U)×t=1

S=1/2(U+at+U)×t

S=1/2(2U+at)×t

S=1/2(2U+at²)

S=Ut+1/2at²

S=1/2(V+U)×t

2s=(V+U)×(V-U) /a

2s×a=(V+U)(V-U)

2as=V²-U²

V²-U²=2as

Hope this is helpful for u :)

Answer 2

Hello friend

___________________________

We can determine the distance covered by the object with time t from the area of the quadrangle DOEB.

DOEB is a trapezium.So we can use the formula for its area.

s = area of trapezium DOEB

s = 1/2 x sum of lengths of parallel sides x distance between the parallel sides

s = 1/2 x (OD + BE) x OE

but,OD = u, BE = v and OE = t

s = 1/2 x (u + v) x t .............( i )

But,we know that:

a = v - u/t

: t = ( v - u)/a ................( ii )

Now,putt the value of t in the equation ( i )

So,

s = 1/2 x ( u + v) x t

s = 1/2 x (u + v ) x (v - u)/a

s = ( u + v ) (v - u )/a

s = (u + v ) ( v - u )/2a

2as = (u + v ) ( v - u ) = v^2 - u^2

v^2 = u^2 + 2as

This equation is known as the Newton's Third equation of motion..

Thanks.

:)