Question

derive s=ut+1/2at^2​

Answer 1

■---■---■---■

FIRST EQUATION :-

Initial Velocity = u

BC = BC + CD = BD + OA

BC = v

OA = u

v = BD + u

Or, BD = v - u ....( 1 )

From the velocity - time graph the accelreation of the object is given by -

$a = \frac{change \: in \: velocity}{time}$

$a = \frac{v - u}{t}$

$a = \frac{BD}{t}$

or, BD = at ....( 2 )

Using equations ( 1 ) and ( 2 ) we get,

v = u + at

SECOND EQUATION :-

s = Area of OABC ( which is a trapezium )

s = Area of the rectangle OADC + Area of the triangle ABD

$OA \times OC + \frac{1}{2} (AD \times BD)$

OA = u

OC = AD = t

BD = at

$s = u \times t + \frac{1}{2} (t \times at)$

Or

$s = ut + \frac{1}{2} {at}^{2}$

■---■---■---■