When the acceleration is constant, the v–t graph is a straight line. That one fact gives the two main equations of motion, and you can read both straight off the graph.
Key idea
For constant acceleration: $v = v_0 + at$ and $x = x_0 + v_0 t + \tfrac12 a t^2$. They hold only while the acceleration stays the same.
Why it works
A straight $v$–$t$ line starting at $v_0$ with slope $a$ has height $v_0 + at$ at time $t$: that is the first equation. The displacement is the area under that line, which splits into a rectangle ($v_0 \times t$) and a triangle on top ($\tfrac12 \times t \times at$). Adding them gives $\Delta x = v_0 t + \tfrac12 a t^2$, the second equation.
- Choose a positive direction and write every quantity with its sign.
- List what you know ($x_0$, $v_0$, $a$, $t$, $v$) and what you want.
- Pick the equation with the unknown and no other missing quantity, then use algebra to solve for it.
Worked examples
Worked example: Final speed
A car at $12\ \text{m/s}$ accelerates at $2.5\ \text{m/s}^2$ for $6\ \text{s}$. How fast is it going?
- $v = v_0 + at = 12 + (2.5)(6)$.
- $v = 27\ \text{m/s}$.
Answer: $27\ \text{m/s}$
Worked example: Distance covered
A train starts from rest and accelerates at $0.8\ \text{m/s}^2$ for $20\ \text{s}$. How far does it go?
- $$\Delta x = v_0 t + \tfrac12 a t^2 = 0 + \tfrac12(0.8)(20)^2.$$
- $\Delta x = 0.4 \times 400 = 160\ \text{m}$.
Answer: $160\ \text{m}$
Worked example: Braking to a stop
A car at $20\ \text{m/s}$ brakes with $a = -5\ \text{m/s}^2$. How far does it go before it stops?
- Time to stop ($v = 0$): $0 = 20 - 5t$, so $t = 4\ \text{s}$.
- $$\Delta x = (20)(4) + \tfrac12(-5)(4)^2 = 80 - 40 = 40\ \text{m}.$$
- After $4\ \text{s}$ the car stays put; the equations don't apply once it has stopped.
Answer: $40\ \text{m}$
Common mistake
Using these equations for a whole trip when the acceleration changes partway, like a car that speeds up and then cruises. Then $a$ is not one number, and plugging the total time into $\tfrac12 a t^2$ gives a wrong answer. Fix: split the trip where the acceleration changes; the final velocity and position of one stage are the starting values of the next.