Question

equations of motion​

Answer 1

Equations Of Motion:-

• $v = u + at$

• $s = ut + \dfrac{1}{2} {at}^{2}$

• ${v}^{2} - {u}^{2} = 2as$

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Additional Information

u = initial velocity

v = final velocity

a = acceleration

t = time

s = displacement or distance

g = acceleration due to gravity

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Some examples

➣

If initial velocity is 0

Then, v = u + at

v = 0 + at

v = at

v = gt

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If initial velocity is zero in this case. Then..

$s = ut + \dfrac{1}{2} {at}^{2}$

$s = 0 \times t + \dfrac{1}{2} {at}^{2}$

$s = 0 + \dfrac{1}{2} {at}^{2}$

$s = \dfrac{1}{2} {at}^{2}$

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If initially the body is at rest in this case...

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

${v}^{2} - 0 = 2as$

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

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Initial velocity is zero means the body is initially at rest.

Answer 2

Answer:

$\\\\1) \: v = u + at \\ where \: \\ v \: = final \: velocity \\ u \: = initial \: velocity \\ a \: = acceleration$

$2) \: {v}^{2} = {u}^{2} + 2aS ...(i)\\\: \: or \: \: {v}^{2} = {u}^{2} + 2gh...(ii)\\ \: \\ if \: acceleration \: is \: given \: we \: use \: equation \: (i) \\ if \: acceleration \: due \: to \: gravity \: is \: given \: we \: use \: equation \: (ii)$

$3) \: S = ut + \frac{1}{2} a {t}^{2} \\ where \: S \: = distance \\ u \: = initial \: velocity \\ a \: = acceleration \\ t \: = time$

Explanation:

Derivation of Equation by Algebra Method:

(1)

We know that,

acceleration= rate of change of velocity

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

Cross multiplying, we get:

$at = v - u$

$v = u + at$

Hence Proved

(2)

We already have v = u + at.

We can write above equation as below

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

$\tt \: distance \: = \: average \: velocity \times time$

and

$\tt \: average \: velocity \: = \frac{initial \: velocity+ final \: velocity}{2}$

$\tt \: average \: velocity = \frac{v + u}{2}$

Therefore,

$\\\tt\: distance \: = \frac{v + u}{2} \times \frac{v - u}{a}$

$\tt \: S = \frac{{v}^{2} - {u}^{2} }{2a}$

Cross multiplying, we get:

$\\\tt \: {v}^{2} - {u}^{2} = 2aS$

If acceleration is not given and acceleration due to gravity is given , then instead of a we substitute the value of g (i.e. 9.8 m/s^2 ) and the value distance is substituted as h.

Hence Proved.

(3)

We know that,

$\\\tt \: distance = average \: velocity \: \times time$

$\tt \: average \: velocity = \frac{v + u}{2}$

$\sf \: S = (\frac{v + u}{2} ) \times t$

We know that,

Substituting the value of v in the above equation,

$\\\tt \: v = u + at$

$\sf \: S = ( \frac{u + at + u}{2} ) \times t$

$\sf \: S = (\frac{2u + at}{2}) \times t$

We can write above equation as below,

$\\\sf \: S =( u + \frac{1}{2} at)t$

Multiplying t , we get:

$\\\sf \: S = ut + \frac{1}{2} a {t}^{2}$

Hence Proved.

Special Cases:

If initial velocity is zero(i.e. body is at rest).

$\\\\\sf\:v\:=\:at$

$\sf\:v^2\:=\:2as\\or\\v^2\:=\:2gh$

$\sf\:S\:=\:\frac{1}{2} a t^2$