Question

Derivations of motion equations

Answer 1

Answer:

The relations between these quantities are known as the equations of motion. In case of uniform acceleration, there are three equations of motion which are also known as the laws of constant acceleration. Hence, these equations are used to derive the components like displacement (s), velocity (initial and final), time (t) and acceleration (a).

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Answer 2

Answer:

Answer:

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$\huge \sf Equations \: of \: Motion$

$\sf Consider \: a \: body \: moving \: with \: initial \: velocity \: 'u' \: changes \: it's \: velocity \: to$

$\sf 'v' \: after \: 't' \: seconds. \: Let \: 's' \: is \: the \: displacement \: and \: 'a' \: is \: the$

$\sf acceleration \: of \: the \: body.$

From the defenition of acceleration

a = Change in velocity / time

= (v- u )/ t

by cross multiplication

. at = v - u

$\sf \implies v = u + at$

2) displacement (s) = average velocity × time

S = (u + v /2) × t

= [u + (u + at )] / 2 ×t

=( 2u + at ) / 2. × t

= 2ut/2 + 1/2 × ${at}^{2}$

$\sf \implies S \: = \: ut \: + \: \frac{1} {2} {at}^{2}$

3). v = u + at

by squaring on both sides

$\sf {v}^{2} \: = \: ({u \: + \: at})^{2}$

= $\sf {u}^{2} \: + \: 2uat \: + \: {a}^{2} {t}^{2}$

$\sf \implies {v}^{2} \: = \: {u}^{2} \: + \: 2as$

Hope it helps you.....