Question

12
Derive graphically (i) position-time relation (ii) velocity - time relation for an object under
uniformly accelerated motion​

Answer 1

We are asked to derive,

⋆ The two equations of motion by using graphical method, we have to derive them 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)

~ Now let us derive velocity-time relation first. Let's do it.

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

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

As we write BD at the place of v-u henceforth,

$\tt \Rightarrow v - u \: = at \\ \\ \tt \Rightarrow v = \: u \: + at \\ \\ {\pmb{\sf{Henceforth, \: derived!}}}$

~ Now let us derive position-time relation.

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

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