Question

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Brainly wizards,

${\Large{\underline{\underline{\bf{\maltese Question \maltese}}}}}$

Derive Newton's third equation of motion . Give a example also related to third equation of motion.

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Answer 1

Answer:

We know that, displacement is the rate of change of position of an object. Mathematically, this can be represented as:

Displacement=(InitialVelocity+FinalVelocity2)×t

hi

Substituting the standard notations, the above equation becomes

s=(u+v2)×t

From the first equation of motion, we know that

v=u+at

Rearranging the above formula, we get

t=v−ua

Substituting the value of t in the displacement formula, we get

s=(v+u2)(v−ua)

s=(v2−u22a)

2as=v2−u2

Rearranging, we get

v2=u2+2as

Through graphical method

From the graph, we can say that

The total distance travelled, s is given by the Area of trapezium OABC.

Hence,

S = ½ (Sum of Parallel Sides) × Height

S=(OA+CB)×OC

Since, OA = u, CB = v, and OC = t

The above equation becomes

S= ½ (u+v) × t

Now, since t = (v – u)/ a

The above equation can be written as:

S= ½ ((u+v) × (v-u))/a

Rearranging the equation, we get

S= ½ (v+u) × (v-u)/a

S = (v2-u2)/2a

Third equation of motion is obtained by solving the above equation:

v^2 = u^2+2aS

Example in the attachment.

Answer 2

$\rm{\underline{\underline{Third \: equation \: of \: motion:–}}}$

Derivation of 3rd equation of motion

→ v² = 4 + 2as

For any object in uniform motion

$\rm→ \: s = \frac{distance}{displacement}$

→ u = initial velocity

→ v = final velocity

→ t = time

→ a = acceleration

From the Ist equation of motion we know that,

→ v = u + at

$\rm→ \frac{v - u}{a} = t$

$\rm→t = \frac{v - u}{a} →1$

also, from 2nd equation we know that,

→ s = it + ½at²

$\rm→ s = u( \frac{v - u}{a} ) + \frac{a \times (v - u)}{2 \times a²}$

= substituting t from equation 1 we get,

$\rm{s = u (\frac{v - u}{a}) +½ \: a( \frac{v - u}{a} )² }$

$\rm → s = \frac{uv-u²}{a} \times 2+ \frac{v² - 2vu + {4}^{2} }{2a}$

$\rm → s = \frac{2uv - 2u²}{2a} + \frac{v²-2vu+4²}{2a}$

$\rm → s = \frac{2uv - 2u²+v²-2vu+u²}{2a}$

$\rm → s = \frac{-u²+v²}{2a}$

$\rm → 2aS \: = \: -u²+v² \\ → v² = u²+2aS$