Derivation of Equation of Motion - 3 Equations of Motion

Derivation of First Equations of Motion:

$$
a=\Delta v / \Delta t
$$

Expand $\Delta v$ to $v-u$ and condense $\Delta t$ to $t$.

$$
a=(v-u) / t
$$
Then solve for $v$ as a function of $t$.

$$
\mathrm{v}=\mathrm{u}+\mathrm{at}
$$

Derivation of Second Equations of Motion:

Start with the definition of average velocity.

$$
v=\Delta s / \Delta t
$$

Expand $\Delta s$ to $s-s_0$ and condense $\Delta t$ to $t$.

$$
v=\left(s-s_0\right) / t
$$
Solve for the position.

$$
\begin{aligned}
& s=s_0+v t \ldots[a] \\
& v=1 / 2(v+u) \ldots \ldots[4]
\end{aligned}
$$
To continue, we'd like to resort to a touch trick referred to as the mean speed theorem or the Merton rule. I prefer the latter since the rule is applied to any amount that changes at an even rate — not simply speed. The Merton rule was 1st revealed in 1335 at Merton faculty, Oxford by land thinker, a man of science, logician, and calculator William Heytesbury (1313–1372). Once the speed of modification of an amount is constant, its average price is halfway between its final and initial values.

Substitute the first equations of motion [1] into equation [4] and simplify with the intent of eliminating v.

a relentless term $\left(\mathrm{s}_0\right)$, followed by a primary order term (ut ), followed by a second-order term ( $1 / 2 \mathrm{at}^2$ ).

The image $s_0$ is commonly thought of because of the initial position. The image $s$ is that the position it slows to later. you'll decide the ultimate position if you need to. The amendment in position ( $\Delta s$ ) is termed the displacement or distance (depending on circumstances) and a few individuals like writing the second equations of motion or derivation of equations of motion like this.

$$
\Delta s=u t+1 / 2 a t^2
$$

Derivation of Third Equations of Motion:

The first 2 equations of the motion formula each describe one kinematic variable as an operation of your time. In essence…

1. Speed is directly proportional to time once acceleration is constant $(v \propto t)$.
2. Displacement is proportional to time square once acceleration is constant $\left(\Delta s \propto t^2\right)$.

3. Displacement is proportional to speed square once acceleration is constant ( $\Delta s \propto v^2$ ). This statement is especially relevant to driving safety. Once you double the speed of an automobile, it takes fourfold a lot of distance to prevent it. Triple the speed and you may want ninefold a lot of distance. This is often an honest rule of thumb to recollect. The abstract introduction is finished.Time to derive the formula derivation of equations of motion. method 1 Combine the primary 2 equations along in a very manner that may eliminate time as a variable. $v=v_0+$ at..... [1]

solve it for time…

$\mathrm{t}=\left(\mathrm{v}-\mathrm{v}_0\right) /$ aand then substitute it into the second equations of motion formula...

$$
s=s_0+v_0 t+1 / 2 a t^2
$$
Substitute the value of $t$ in equation [2] and solving we get

$$
\begin{aligned}
& 2 a\left(s-s_0\right)=v^2-v_0^2 \\
& v^2=v_0^2+2 a\left(s-s_0\right)
\end{aligned}
$$
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