Deriving the suvat Formulas

A velocity-time (v-t) graph for constant acceleration is always a straight line because the rate of change of velocity is uniform.

Deriving $v = u + at$: The gradient of a velocity-time graph represents acceleration. By calculating the gradient as $\frac{\text{change in velocity}}{\text{time}}$, we get $a = \frac{v - u}{t}$, which rearranges to the first equation.

Deriving $s = \frac{1}{2}(u + v)t$: The area under a velocity-time graph represents displacement. For a constant acceleration graph, this area forms a trapezium with parallel sides of lengths $u$ and $v$ and a height of $t$.

Deriving $s = ut + \frac{1}{2}at^2$: This is found by splitting the area under the graph into a rectangle (area $ut$) and a triangle (area $\frac{1}{2} \times t \times (v - u)$). Substituting $(v - u) = at$ into the triangle area yields the final form.

Since acceleration is the derivative of velocity with respect to time ($a = \frac{dv}{dt}$), velocity can be derived by integrating acceleration: $v = \int a \, dt = at + C$.

By applying the initial condition that at $t = 0$, velocity $v = u$, we find that the constant of integration $C = u$, resulting in $v = u + at$.

Similarly, displacement is the integral of velocity: $s = \int v \, dt = \int (u + at) \, dt = ut + \frac{1}{2}at^2 + C$.

Given that at $t = 0$, displacement $s = 0$ (relative to the start), the constant $C$ becomes $0$, yielding the standard displacement formula.

The equation $v^2 = u^2 + 2as$ is derived by eliminating the time variable ($t$) from the other equations.

By rearranging $v = u + at$ to $t = \frac{v - u}{a}$ and substituting this into $s = \frac{1}{2}(u + v)t$, we obtain $s = \frac{1}{2}(u + v)(\frac{v - u}{a})$.

Expanding the numerator (a difference of two squares) gives $s = \frac{v^2 - u^2}{2a}$, which rearranges to the final squared velocity formula.

Check the Units: Always ensure that $a$ is in $m/s^2$, $v$ and $u$ in $m/s$, and $t$ in $s$ before attempting a derivation or calculation.

Directional Consistency: When deriving or using these formulas, define a positive direction (e.g., upwards or to the right) and stick to it for $s, u, v,$ and $a$.

Identify the 'Missing' Variable: Each of the five equations lacks exactly one of the five variables. In exams, identify which variable is not mentioned and choose the formula that excludes it.

Verification: If asked to derive $s = ut + \frac{1}{2}at^2$, remember to explicitly state that the area under the v-t graph is being calculated.