Question

Derive equations of motion

Answer 1

Derivation of the Equations of Motion

v = u + at

Let us begin with the first equation, v=u+at. This equation only talks about the acceleration, time, the initial and the final velocity. Let us assume a body that has a mass “m” and 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

Therefore, Acceleration = (Final Velocity-Initial Velocity) / Time Taken

Hence, a = v-u /t or at = v-u

Therefore, we have: v = u + at

v² = u² + 2as

We have, v = u + at. Hence, we can write t = (v-u)/a

Also, we know that, Distance = average velocity × Time

Therefore, for constant acceleration we can write: Average velocity = (final velocity + initial velocty)/2 = (v+u)/2

Hence, Distance (s) = [(v+u)/2] × [(v-u)/a]

or s = (v² – u²)/2a

or 2as = v² – u²

or v² = u² + 2as

s = ut + ½at²

Let the distance be “s”. We know that

Distance = Average velocity × Time. Also, Average velocity = (u+v)/2

Therefore, Distance (s) = (u+v)/2 × t

Also, from v = u + at, we have:

s = (u+u+at)/2 × t = (2u+at)/2 × t

s = (2ut+at²)/2 = 2ut/2 + at²/2

or s = ut +½ at²

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.....