Let's derive the formulars of linear acceleration

Morning Abba — let's get this done. Let's derive the four main kinematic equations for linear motion with constant acceleration. Assumptions: • Initial velocity: u • Final velocity: v • Acceleration: a (constant) • Time: t • Displacement: s Step 1: Derive the first equation (v = u + at) Acceleration is defined as the rate of change of velocity. a = change in velocitytime taken = (v - u)/(t) Rearranging this equation to solve for v: at = v - u v = u + at Step 2: Derive the second equation (s = ut + (1)/(2)at^2) Displacement is the area under the velocity-time graph. For constant acceleration, the graph is a straight line. The area can be split into a rectangle and a triangle. Area of rectangle = u × t = ut Area of triangle = (1)/(2) × base × height = (1)/(2) × t × (v - u) From Step 1, we know v - u = at. Substitute this into the triangle area: Area of triangle = (1)/(2) × t × (at) = (1)/(2)at^2 Total displacement s is the sum of these areas: s = ut + (1)/(2)at^2 Step 3: Derive the third equation (v^2 = u^2 + 2as) From Step 1, we have t = (v - u)/(a). Substitute this expression for t into the equation from Step 2 (s = ut + (1)/(2)at^2): s = u((v - u)/(a)) + (1)/(2)a((v - u)/(a))^2 s = (uv - u^2)/(a) + (1)/(2)a((v - u)^2)/(a^2) s = (uv - u^2)/(a) + ((v - u)^2)/(2a) Multiply by 2a to clear the denominators: 2as = 2(uv - u^2) + (v - u)^2 2as = 2uv - 2u^2 + v^2 - 2uv + u^2 2as = v^2 - u^2 Rearranging to solve for v^2: v^2 = u^2 + 2as Step 4: Derive the fourth equation (s = ((u+v))/(2)t) Displacement can also be calculated as the average velocity multiplied by time. Average velocity = (u + v)/(2) s = ((u + v)/(2))t These are the four main formulas for linear acceleration: 1. v = u + at 2. s = ut + (1)/(2)at^2 3. v^2 = u^2 + 2as 4. s = ((u+v))/(2)t Send me the next one 📸