Question

Derive the 3 equations of motion and second law 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



Now second law of motion:
Second law: In an inertial reference frame, the vector sum of the forces F on an object is equal to the mass m of that object multiplied by the acceleration a of the object: F = ma. (It is assumed here that the mass m is constant – see below.)

Answer 2

There are three equations of motion which are also known as the laws of constant acceleration.
These equations are use to derive like
= Displacement(s)
=velocity (initial and final)
=Time (t)
= Acceleration (a)

First equation of motion
(v=u+at)
a=v-u/t
v-u=at
v=u+at

second equation of motion
(s=ut+at square)
s=(u+v)/2×t
v=u+at(from first equation of motion)
s=(u+u+at)/2×t
s=(2u+at)/2×t
2ut+at (square )/2
2ut+at square
ut+1\2at square
s=ut+1\2at square

third equation of motion
( v square=u square+ 2as)
t=v-u/a
s=u(v-u)/a+1\2a[v-u/a] square