**Derivation of the Third Equation of Motion**

The third equation of motion relates the final velocity of an object to its initial velocity, acceleration, and displacement. It can be mathematically derived using the principles of kinematics and basic calculus. Let's break down the derivation into several steps:

**Step 1: Defining the Variables**
- Let's assume that an object starts from rest with an initial velocity of u.
- The object undergoes constant acceleration a for a time interval t.
- During this time, the object covers a displacement s.

**Step 2: Defining the Relationship**
- The goal is to find a mathematical relationship between u, v (final velocity), a, and s.
- The first equation of motion tells us that v = u + at.
- Integrating this equation will help us derive the desired relationship.

**Step 3: Integrating the Equation**
- Integrating the equation v = u + at with respect to time t yields:
∫dv = ∫(u + at) dt
- Integrating both sides gives us:
v = ut + (1/2)at^2 + C

**Step 4: Determining the Constant of Integration**
- The constant of integration (C) represents the initial condition when t = 0.
- At t = 0, v = u, so we can substitute these values into the equation:
u = u(0) + (1/2)a(0)^2 + C
- Since u(0) = 0 and a(0) = 0 (object starts from rest), the equation simplifies to:
u = C

**Step 5: Final Equation**
- Substituting the value of C into the previously integrated equation, we have:
v = ut + (1/2)at^2
- Rearranging the equation gives:
v = u + (1/2)at^2

**Conclusion**
By following the steps outlined above, we have mathematically derived the third equation of motion: v = u + (1/2)at^2. This equation relates the final velocity (v) of an object to its initial velocity (u), acceleration (a), and the time interval (t) during which the acceleration occurs.

Consider a body having initial velocity 'u'. Suppose it is subjected to a uniform acceleration 'a' so that after time 't' it's final velocity becomes 'v' after travelling distance 's' Now from second equation of motion we have S=ut+1/2at^2......... (1)And from first equation of motion we have v=u +atThis can be rearranged and written as:at=v-ut=v-u/as Now,putting the value of 't' in equation (1), we get:S=u(v-u) /a +1/2a[v-u/a]^2S=uv-u^2/a+a(v^2+u^2-2uv)/2a^2..[bcz (v-u) ^2=v^2+u^2-2 uv] S=uv-u^2/a+v^2+u^2-2uv/2aS=2uv-2u^2+v^2+u^2-2uv/2a2as=v^2-u^2V^2=u^2 +2asHope it help u I hv another solution too if u want I can give u :)