Question

Derivation of 3 equations of motion

Answer 1

1.First Equation of Motion-:

We know ,

acceleration (a) =ChangeinvelocityTimetakenChangeinvelocityTimetaken

a=Finalvelocity−InitialvelocityTimetakenFinalvelocity−InitialvelocityTimetaken

or => a= (v−u)t(v−u)t

or => at= v-u

or => v= u+ at —————(i)

2. Second Equation of Motion -:

We know ,

=> Distance travelled = Average Velocity × Time

=> Distance travelled = InitialVelocity+FinalVelocity2×TimeInitialVelocity+FinalVelocity2×Time

=> s=v+u2×tv+u2×t

From equation (i) => v=u+at

∴ s = u+(u+at)2×tu+(u+at)2×t

or s = (2u+at)2×t(2u+at)2×t

or s= ut + (1/2 • at²) —————(ii)

3.Third equation of motion -:

=> Distance travelled = Average Velocity × Time

=> Distance travelled =InitialVelocity+FinalVelocity2×TimeInitialVelocity+FinalVelocity2×Time

=> s= v+u2×tv+u2×t

From equation (i) => v=u+at

or t=v−uav−ua

∴ s= u+v2×u+v2×v−uav−ua = v²−u²2av²−u²2a

or v²-u² = 2as

or v²= u²+2as ——————(iii)

Where ,

u =initial velocity

v=final velocity

t= time

s= distance

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.....