Question

pleade hep me to solve this ​

Answer 1

We are asked to derive,

⋆ The one equation of motion by using graphical method, we have to derive it graphically. Let's derive!

Firstly, before deriving let us know that what are equations of motion.

⋆ There are three equations of motion. The equations of motion are named as

◆ Velocity time relationship

◆ Position time relationship

◆ Position velocity relationship

⋆ Therefore, velocity time relationship, Position time relationship and Position velocity relationship are the three equations of motion respectively.

◆ Velocity time relationship

That's the equation,

• ${\small{\underline{\boxed{\sf{v \: = u \: + at}}}}}$

◆ Position time relationship

That's the equation,

• ${\small{\underline{\boxed{\sf{s \: = ut \: + \dfrac{1}{2} \times at^2}}}}}$

◆ Position velocity relationship

That's the equation,

• ${\small{\underline{\boxed{\sf{2as \: = v^2 - u^2}}}}}$

(Where, v denotes final velocity , u denotes initial velocity , a denotes acceleration , t denotes time , s denotes displacement or distance)

~ Firstly according to the graph,

⇢ AO = DC = u (Initial velocity)

⇢ AD = OC = t (Time)

⇢ EO = BC = v (Final velocity)

Firstly let, the object is travelling a distance in time under uniform acceleration. Now according to the graph we are able to see that inside the graph the obtained area enclose within OABC(trapezium) under velocity time graph AB. Therefore, distance travelled by an object can be given by

$\tt \Rightarrow Distance \: = Area \: enclosed \\ \\ \tt \Rightarrow s \: = Area \: enclosed \\ \\ \tt \Rightarrow s \: = OABC \\ \\ \tt \Rightarrow s \: = Area \: of \: rectangle \: + Area \: of \: \triangle \\ \\ \tt \Rightarrow s \: = Length \times Breadth + \dfrac{1}{2} \times Base \times Height \\ \\ \tt \Rightarrow s \: = AO \times AD + \dfrac{1}{2} \times AD \times BD \\ \\ \tt \Rightarrow s \: = u \times t + \dfrac{1}{2} \times t \times BD \\ \\ \tt \Rightarrow s \: = u \times t + \dfrac{1}{2} \times t \times at \\ \\ \tt \Rightarrow s \: = ut + \dfrac{1}{2} \times at^2 \\ \\ {\pmb{\sf{Henceforth, \: derived!}}}$

How the value of BD came?

Firstly we can write BC as BD + DC. Now as BD have the velocity position and DC have the time position. Henceforth, we already know that

$\tt \Rightarrow Acceleration \: = \dfrac{Change \: in \: velocity}{Time} \\ \\ \tt \Rightarrow a \: = \dfrac{Change \: in \: velocity}{Time} \\ \\ \tt \Rightarrow a \: = \dfrac{v-u}{t} \\ \\ \tt \Rightarrow a \: = \dfrac{BD}{t} \\ \\ \tt \Rightarrow at \: = BD$