Where the SUVAT Equations Come From

Build the four equations from that one line, one at a time. Each is just a small step from the line or from the equation before it.

Equation 1 — from the gradient

1. The line goes up from $u$ at the start to $v$ at time $t$.
2. Gradient is the acceleration: $a = \dfrac{v-u}{t}$.
3. Multiply both sides by $t$: $at = v - u$.
4. Add $u$ to both sides: $v = u + at$.

Equation 2 — area as a trapezium

1. The displacement is the area under the line.
2. That area is a trapezium: the two upright sides are $u$ and $v$, and the width is $t$.
3. Area of a trapezium = average of the two sides × width.
4. The average of $u$ and $v$ is $\tfrac{1}{2}(u+v)$, so $s = \tfrac{1}{2}(u+v)\,t$.

Equation 3 — split the same area

1. Cut the same area into a rectangle with a triangle on top.
2. Rectangle: height $u$, width $t$, so its area is $ut$.
3. Triangle: width $t$, height $v-u$, which is $at$ from Equation 1.
4. Triangle area $= \tfrac{1}{2}\times t \times at = \tfrac{1}{2}at^2$.
5. Add the two parts: $s = ut + \tfrac{1}{2}at^2$.

Equation 4 — remove t

1. From Equation 1, make $t$ the subject: $t = \dfrac{v-u}{a}$.
2. Put this into Equation 2: $s = \tfrac{1}{2}(u+v)\times\dfrac{v-u}{a}$.
3. The top becomes $(u+v)(v-u) = v^2 - u^2$, so $s = \dfrac{v^2 - u^2}{2a}$.
4. Rearrange: $v^2 = u^2 + 2as$.

• Each equation uses four of the five letters and leaves one out. That fact becomes the method on the next page.
• Only $s = ut + \tfrac{1}{2}at^2$ and $v^2 = u^2 + 2as$ are printed on the exam data sheet. Know the other two from memory.
• A show-that question can ask you to build any of these from the graph, so learn the gradient-and-area reasoning, not just the four results.