Question

Derive: S= ut +1/2at2 with the help of graphical method.​

Answer 1

To prove :

$\bullet \: \sf{s =ut + \dfrac{1}{2} {at}^{2} }$

Prove :

$\sf{ \implies \: s = Area \: of \: trapezium \: OABCD } \\ \\ \sf{ \implies s = Area \: of \: reactangle + Area \: of \: triangle} \\ \\ \implies\sf{s = OD \times CD + \dfrac{1}{2} \times AC \times BC} \\ \\ \implies\sf{s = t \times u + \dfrac{1}{2} \times t \times (v - u)} \\ \\ \implies\sf{s = ut + \frac{1}{2} \times t \times at } \: \: \: \: \: \sf \red{ \bigg(v - u = at \bigg) }\\ \\ \implies\sf{s = ut = \dfrac{1}{2} {at}^{2} }$

_______...

There are three equations of motion :

• v = u + at
• s = ut + ½ at²
• v² = u² + 2as

• v stands for Final velocity

• u stand for Initial velocity

• s stands for Distance

• a stands for Acceleration

• t stands for Time

Answer 2

Answer:

To prove :

\bullet \: \sf{s =ut + \dfrac{1}{2} {at}^{2} }∙s=ut+

2

1

at

2

Prove :

\begin{gathered}\sf{ \implies \: s = area \: of \: trapezium \: oab cd } \\ \\ \sf{ \implies s = Area \: of \: reactangle + Area \: of \: triangle} \\ \\ \implies\sf{s = OD \times CD + \dfrac{1}{2} \times AC \times BC} \\ \\ \implies\sf{s = t \times u + \dfrac{1}{2} \times t \times (v - u)} \\ \\ \implies\sf{s = ut + \frac{1}{2} \times t \times at } \: \: \: \: \: \sf \red{ \bigg(v - u = at \bigg) }\\ \\ \implies\sf{s = ut = \dfrac{1}{2} {at}^{2} }\end{gathered}

⟹s=areaoftrapeziumoabcd

⟹s=Areaofreactangle+Areaoftriangle

⟹s=OD×CD+

2

1

×AC×BC

⟹s=t×u+

2

1

×t×(v−u)

⟹s=ut+

2

1

×t×at(v−u=at)

⟹s=ut=

2

1

at

2