A velocity-time ($v$–$t$) graph looks like the x–t graphs you know, but it answers different questions. Its height is the velocity, and two features carry the rest of the story: its slope and the area under it.
Key idea
On a $v$–$t$ graph, the slope is the Acceleration and Its Sign and the area between the graph and the time axis is the displacement (area below the axis counts as negative).
Why it works
Slope: rise over run is $\Delta v / \Delta t$, which is the definition of acceleration. Area: at a steady velocity, displacement is $v \times \Delta t$, which is the area of a rectangle of height $v$ and width $\Delta t$. When $v$ changes, slice the time into thin strips; each strip is nearly a rectangle, so adding the strips adds up the little displacements. Shapes from Area of a Trapezoid do the adding for you when the pieces are straight.
Worked examples
Worked example: Slope gives acceleration
A cart's velocity rises in a straight line from $2\ \text{m/s}$ at $t = 0$ to $8\ \text{m/s}$ at $t = 3\ \text{s}$. Find its acceleration.
- Slope $= \dfrac{\Delta v}{\Delta t} = \dfrac{8 - 2}{3 - 0}$.
- $a = 2\ \text{m/s}^2$.
Answer: $2\ \text{m/s}^2$
Worked example: Area gives displacement
For that same cart (from $2$ to $8\ \text{m/s}$ in $3\ \text{s}$), how far does it go?
- The area under the line is a trapezoid with parallel sides $2$ and $8\ \text{m/s}$ and width $3\ \text{s}$.
- Area $= \tfrac12(2 + 8)(3) = 15$. The units are $\dfrac{\text{m}}{\text{s}} \cdot \text{s} = \text{m}$.
Answer: $\Delta x = 15\ \text{m}$
Worked example: Area below the axis
A cart moves at $+4\ \text{m/s}$ for $3\ \text{s}$, then at $-2\ \text{m/s}$ for $2\ \text{s}$. Find its displacement and the distance it travels.
- Area above the axis: $4 \times 3 = 12\ \text{m}$. Area below: $-2 \times 2 = -4\ \text{m}$.
- Displacement $= 12 + (-4) = 8\ \text{m}$. Distance $= 12 + 4 = 16\ \text{m}$.
Answer: $\Delta x = 8\ \text{m}$; distance $16\ \text{m}$
Common mistake
Reading a $v$–$t$ graph as if it showed position, so a downward slope "means moving backward". A downward slope means the velocity is decreasing; the cart still moves forward as long as the graph is above the axis. Check: direction of motion comes from the sign of $v$ (above or below the axis); the slope only says whether $v$ is going up or down.