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.