Question

derive first equation of motion

Answer 1

Consider a body of mass “m” having initial velocity “u”.Let after time “t” its final velocity becomes “v” due to uniform acceleration “a”.

Now we know that:

Acceleration = change in velocity/Time taken

=> Acceleration = Final velocity-Initial velocity / time taken

=> a = v-u /t

=>at = v-u

or v = u + at

This is the first equation of motion.

Answer 2

Answer:

Answer:

$\sf Hlo \: mate \: here's \: your \: answer \: Hope \: it \: helps!$

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

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

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

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

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.....