Question

can someone please explain me the derivations of equations of the motion...

Answer 1

first equation of motion, second equation of motion and third equation of motion.

Answer 2

$\textbf{ Hello Mate! }$

1. As we know that slope of graph =
( y2 - y1 ) / ( x2 - x1 )

Where, velocity time graph slope gives acceleration.

a = ( y2 - y1 ) / ( x2 - x1 )

a = ( v - u ) / ( t - 0 )

at = v - u

$\textbf{ v = u + at }$

2. The area enclosed under slope is distance travelled or displacement.

Area as given in image = Area of [] + Area of ∆

s = ( l × b ) + ( 1 / 2 × b × h )

s = ( t × u ) + [ 1 / 2 × t × ( v - u )]

[ Here, v - u = at ]

s = ut + 1/2 × t × at

$\textbf{s = ut + 1/2 ( at^2 )}$

3. Now, can't we see area as trapezium?

Area of trapezium = 1/2 × h × ( sum of parallel sides )

s = 1 / 2 × t × ( v + u )

Now, t = ( v - u ) / a

s = 1 / 2 × ( v - u ) / a × ( v + u )

s = ( v² - u² ) / 2a

$\textbf{ 2as = v^2- u^2 }$

Hope it helps