Question

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Derive the formula of third equation of motion i.e v² - u² = 2as

Answer 1

Answer:

For getting the third law of motion equation, we use both of the equations of motion.

S = ut + 1/2^at²__________(1)

v = u + at _____________ (2)

t = (v – u)/a

Substitute the value of t in equation (1)

Substitute the value of t in equation (1)S = ut + 1/2^at²

s= u(v – u)/a)+1/2^a(v-u)²

2aS = 2u(v−u) + (v−u)²

2aS = 2uv − 2u²+v²-2uv+u²

2aS= v²-u²

v²= u²+2aS..

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Explanation:

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Answer 2

We will use both of the equations of motion to reach the third equation of motion. This will require a bit of algebra.

$S = ut + \frac{1}{2} \: at {}^{2} \: and \: V = u + ut, \\ include \: the \: time \: variant \: t.$

There will be some situations when we do not have any information about time and so it would be a good idea to derive an equation that does not have a t term.

To do this, we rearrange our first equation to get

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

and use this to replace t wherever it appears in the second equation. So,

$S = ut + \frac{1}{2} \: at {}^{2} \: becomes,$

$S = u (\: \frac{v - u}{a} ) + \frac{1}{2} \: a( \frac{v - u}{a} ) {}^{2}$

$= > 2aS = 2u(v - u) + (v - u) {}^{2}$

$= > 2aS = 2uv - 2u {}^{2} - v {}^{2} - 2uv - u {}^{2}$

$= > 2aS = v {}^{2} - u {}^{2}$

$\fbox\pink{v² - u² = 2as}$