**Derivations of Equations of Motions by Calculus Method**

Let a particle start moving with velocity u at time t = 0 along a straight line. The particle has a constant acceleration a. Let at t = t, its velocity becomes v and it covers a displacement of s during this time.

v = ∫adt + C₁ ⇒ v = at + C₁

At t = 0, v = u this gives C₁ = u

Putting C₁, we get v = u + at … (1)

x = ∫v dt + C₂ = ∫(u + at) dt + C₂

x = ∫u dt + ∫at dt + C₂ = ut + ½ at² + C₂

At t = 0, x = 0 this gives C₂ = 0

Putting C₂, we get x = ut + ½ at² … (2)

\(v\frac{dv}{dx}=a\),

∫ v dv = ∫ a dx + C₃ \(\Rightarrow \frac{{{v}^{2}}}{2}=ax+{{C}_{3}}\),

At x = 0, v = u this gives \({{C}_{3}}=\frac{{{u}^{2}}}{2}\),

\(\frac{{{v}^{2}}}{2}=ax+\frac{{{u}^{2}}}{2}\),

v² = u² + 2ax.