A speedometer shows your velocity right now, not over a trip. On a curved position-time graph the slope keeps changing, so we need the slope at a single point.
Key idea
The instantaneous velocity at time $t$ is the slope of the tangent line to the $x$–$t$ graph at $t$: the line that just touches the curve there and runs in the same direction.
Why it works
Take the average velocity over a short interval around $t$, then a shorter one, then shorter still. On a smooth curve, zooming in makes the curve look straighter and straighter, and those average-velocity lines settle down to one line: the tangent. Its slope is the velocity at that instant. In calculus this is the derivative, but the idea only needs slopes.
Worked examples
Worked example: From a tangent line
The tangent to an $x$–$t$ curve at $t = 2\ \text{s}$ passes through $(1\ \text{s}, 4\ \text{m})$ and $(3\ \text{s}, 8\ \text{m})$. Find the velocity at $t = 2\ \text{s}$.
- Slope $= \dfrac{8 - 4}{3 - 1} = \dfrac{4\ \text{m}}{2\ \text{s}}$.
- $v = 2\ \text{m/s}$.
Answer: $2\ \text{m/s}$
Worked example: From a short interval
A sensor reads $x = 5.39\ \text{m}$ at $t = 2.9\ \text{s}$ and $x = 5.79\ \text{m}$ at $t = 3.1\ \text{s}$. Estimate the velocity at $t = 3\ \text{s}$.
- Over this short interval, $v \approx \dfrac{5.79 - 5.39}{3.1 - 2.9} = \dfrac{0.40}{0.2}$.
- $v \approx 2.0\ \text{m/s}$. A shorter interval would give an even better estimate.
Answer: about $2.0\ \text{m/s}$
Common mistake
Thinking an object at $x = 0$ must have zero velocity. Position and velocity are different questions: a car passing the origin at highway speed is at $x = 0$ with a large velocity, and a ball at the top of its rise has zero velocity far from the origin. Check: velocity comes from the slope of the graph, never from its height.