Question

Derive the third equation of motion by graphical method.

Answer 1

Answer:hope its help u

Explanation:

Answer 2

The question is incomplete, below you will find the missing graph

Concept:

Third equation of motion is the distance covered by the object moving with uniform acceleration is given by the area of trapezium.

Given:

the initial velocity be u, and the final velocity be v and let the object moving with uniform acceleration.

Find:

We will derive the third equation of motion by graphical method.

Solution:

Let the initial velocity of object be “u” and the final velocity becomes “v” of the object when it reaches at point B after time “t”.

Let the object moving with uniform acceleration “a”.

We know the formula,

Area of trapezium = ½ × (sum of parallel sides × distance between them)

∴ Distance(s) = $\frac{1}{2}$ × ( OD+BE ) × DE

                      = $\frac{1}{2}$ × (u + v ) × t --------1

   As we know that,  a = $\frac{v-u}{t}$

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

 putting the value of "t" in equation 1

                  s = $\frac{1}{2}$ × (u + v) × $\frac{v-u}{a}$

                      = $\frac{1}{2a}$ × (v + u) ×(v - u)

                   2as = $v^{2}$ - $u^{2}$

                 ∴ $v^{2}$ = $u^{2}$ + 2as

Hence the third equation of motion is derived by graphical method.

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