Question

Derive 3rd equation of motion by graphical method

Answer 1

See the following pics and draw a graph for it

Answer 2

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$\huge\bold\green{ANSWER:}$

✿ Derivation of 3rd Equation of motion [Position-Velocity Relation]- graphically:-

2as= v² - u²

As per the graph,

OA= u= initial velocity

BC= v= final velocity

OC= t= time taken

Now, let the body covers a displacement of s in time t...

From the graph,

Displacement= s= Area under graph AB

So, s= Area of the trapezium OABC

We know that,

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

Similarly,

Area of OABC= 1/2× (OA+BC)× OC [Now, substitute all the values from the graph in the expression..]

→s= (u+v)×t/2 [Consider it as the equation-1]

From the first equation of motion, we have→ v= u+at or t= v-u/a [Consider it as the equation- 2]

On substituting equation 2 in equation 1, we get,

→s= (v+u)(v-u)/2a

→s= v² - u²/ 2a

→2as= v² - u² [3rd equation of motion]

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Hope it helps...:-)

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