You can now read slopes on an x–t graph and slopes and areas on a v–t graph. Put those together and any one motion graph lets you sketch the other two. This is one of the most useful skills in mechanics, and one of the most often missed.
Key idea
Going down the chain $x \to v \to a$, take slopes: the value of each graph is the slope of the one above it. Going up $a \to v \to x$, add up areas, starting from the known initial value.
Why it works
Velocity is the rate of change of position, and acceleration is the rate of change of velocity. A rate of change is a slope. Undoing a slope means adding up small changes, and adding up a rate times small time steps is an area. So the three graphs are linked by the same two moves every time.
| x–t graph | v–t graph | a–t graph |
|---|---|---|
| straight, sloping up | flat, above the axis | zero |
| flat | zero | zero |
| curving upward (getting steeper) | sloping up | flat, above the axis |
| curving downward | sloping down | flat, below the axis |
Worked examples
Worked example: From x to v
An $x$–$t$ graph is a straight line rising from $x = 0$ to $x = 6\ \text{m}$ over $0$–$3\ \text{s}$, then flat until $t = 5\ \text{s}$. Sketch the $v$–$t$ graph.
- First piece: slope $= \dfrac{6}{3} = 2\ \text{m/s}$, so $v$ is a flat line at $2\ \text{m/s}$ from $0$ to $3\ \text{s}$.
- Second piece: slope $0$, so $v$ is a flat line on the axis from $3$ to $5\ \text{s}$.
- The $v$ graph is a step down, not a copy of the ramp-then-flat shape.
Answer: $v = 2\ \text{m/s}$ for $0$–$3\ \text{s}$, then $v = 0$.
Worked example: From v to a
A $v$–$t$ graph rises from $0$ to $6\ \text{m/s}$ during $0$–$2\ \text{s}$, stays at $6\ \text{m/s}$ until $5\ \text{s}$, then falls to $0$ at $8\ \text{s}$. Sketch the $a$–$t$ graph.
- Slopes: $\dfrac{6}{2} = 3\ \text{m/s}^2$, then $0$, then $\dfrac{-6}{3} = -2\ \text{m/s}^2$.
- The $a$ graph is three flat pieces: $+3$, $0$, $-2\ \text{m/s}^2$.
Answer: $a = 3$, then $0$, then $-2\ \text{m/s}^2$.
Worked example: From v up to x
Starting at $x = 0$, a cart has $v = 4\ \text{m/s}$ for $2\ \text{s}$, then $v = 0$ for $2\ \text{s}$, then $v = -2\ \text{m/s}$ for $3\ \text{s}$. Describe the $x$–$t$ graph.
- Each piece changes $x$ by its area: $+8\ \text{m}$, then $0$, then $-6\ \text{m}$.
- So $x$ climbs in a straight line to $8\ \text{m}$, stays flat, then falls in a straight line to $2\ \text{m}$.
Answer: Up to $8\ \text{m}$, flat, then down to $2\ \text{m}$.
Common mistake
Copying the shape of one graph onto the next. A ramp on the $x$ graph becomes a flat line on the $v$ graph, and a curve on the $x$ graph becomes a sloped line on the $v$ graph. Check each piece by asking "what is the slope here?" (going down) or "what area builds up here?" (going up), never "what does it look like?"