Let's derive the formulars of linear acceleration

Acceleration is defined as the rate of change of velocity.

$$a = \frac{\text{change in velocity}}{\text{time taken}} = \frac{v - u}{t}$$

Rearranging this equation to solve for $v$:

$$at = v - u$$ $$v = u + at$$

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 = $u \times t = ut$
Area of triangle = $\frac{1}{2} \times \text{base} \times \text{height} = \frac{1}{2} \times t \times (v - u)$
From Step 1, we know $v - u = at$. Substitute this into the triangle area:
Area of triangle = $\frac{1}{2} \times t \times (at) = \frac{1}{2}at^2$
Total displacement $s$ is the sum of these areas:

$$s = ut + \frac{1}{2}at^2$$

From Step 1, we have $t = \frac{v - u}{a}$.
Substitute this expression for $t$ into the equation from Step 2 ($s = ut + \frac{1}{2}at^2$):

$$s = u\left(\frac{v - u}{a}\right) + \frac{1}{2}a\left(\frac{v - u}{a}\right)^2$$ $$s = \frac{uv - u^2}{a} + \frac{1}{2}a\frac{(v - u)^2}{a^2}$$ $$s = \frac{uv - u^2}{a} + \frac{(v - u)^2}{2a}$$

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 $v^2$:

$$v^2 = u^2 + 2as$$

Displacement can also be calculated as the average velocity multiplied by time.
Average velocity = $\frac{u + v}{2}$

$$s = \left(\frac{u + v}{2}\right)t$$

These are the four main formulas for linear acceleration:

1. $\boxed{v = u + at}$
2. $\boxed{s = ut + \frac{1}{2}at^2}$
3. $\boxed{v^2 = u^2 + 2as}$
4. $\boxed{s = \frac{(u+v)}{2}t}$