Morning MR — let's get this done. To derive the equation $S = Vt + \frac{1}{2}at^2$, we start with the definitions of acceleration and average velocity for motion under constant acceleration. Step 1: Define acceleration. Acceleration ($a$) is the rate of change of velocity. If $u$ is the initial velocity and $v$ is the final velocity over a time $t$, then: $$a = \frac{v - u}{t}$$ Rearranging this equation to solve for the final velocity $v$: $$v = u + at$$ Step 2: Define average velocity. For motion with constant acceleration, the average velocity ($\bar{v}$) is the arithmetic mean of the initial and final velocities: $$\bar{v} = \frac{u + v}{2}$$ Step 3: Substitute the expression for $v$ from Step 1 into the average velocity equation from Step 2. $$\bar{v} = \frac{u + (u + at)}{2}$$ $$\bar{v} = \frac{2u + at}{2}$$ $$\bar{v} = u + \frac{1}{2}at$$ Step 4: Relate displacement to average velocity. Displacement ($S$) is the product of average velocity and time: $$S = \bar{v}t$$ Step 5: Substitute the expression for $\bar{v}$ from Step 3 into the displacement equation from Step 4. $$S = \left(u + \frac{1}{2}at\right)t$$ $$S = ut + \frac{1}{2}at^2$$ Step 6: Match the notation. The question uses $V$ for initial velocity instead of $u$. Replacing $u$ with $V$: $$S = Vt + \frac{1}{2}at^2$$ This is the derived equation. The derived equation is: $$\boxed{S = Vt + \frac{1}{2}at^2}$$ Send me the next one 📸