Coaching Centre
Average and instantaneous velocity
Average velocity as the gradient of a chord on a position-time graph, instantaneous velocity as the gradient of a tangent, the limit as the time interval shrinks, and point-to-point speed cameras.
On the Hume Highway, point-to-point cameras photograph every truck at one site and again at another, many kilometres away. A truck can sit at the speed limit as it passes each camera and still be fined. The cameras don't care how fast it was going at the cameras. They measure something else entirely.
Average velocity tells you about a whole interval of time. Instantaneous velocity is the velocity at one moment, the number a speedometer shows (with a direction attached). This page shows how both come from a position–time graph, and how one turns into the other.
Average velocity over an interval
From 1A, $v_{\text{avg}} = \dfrac{\Delta r}{\Delta t}$. On a position–time graph, $\Delta r$ is the rise and $\Delta t$ is the run between two points. So:
Average velocity over an interval = gradient of the chord (the straight line joining the two points on the graph).
A truck passes camera 1 at 10:02:00 and camera 2, 18.0 km further along the highway, at 10:12:00. The limit for heavy vehicles is 100 km/h. Should the truck be fined?
Instantaneous velocity
If the velocity is changing, the average over a long interval hides what happens inside it. Make the interval shorter and the chord gets closer to the curve. In the limit, as $\Delta t \to 0$, the chord becomes the tangent, the straight line that just touches the curve at one point:
$$v = \lim_{\Delta t \to 0} \frac{\Delta r}{\Delta t}$$Instantaneous velocity = gradient of the tangent to the position–time graph at that instant. (If you do Mathematics Advanced, this is the derivative $\frac{dx}{dt}$.)
For uniform motion the position–time graph is a straight line, so every chord and every tangent has the same gradient: average and instantaneous velocity are equal. They only differ when the velocity is changing.
Drawing a tangent in the exam
- Mark the point on the curve at the time you want.
- Lay a ruler so it touches the curve at that point only, with the curve bending away equally on both sides. Draw a long line.
- Pick two points on your line far apart (not on the curve) and read their coordinates.
- Gradient $= \dfrac{\text{rise}}{\text{run}}$, with units (m ÷ s = m s⁻¹). Keep the sign: a downward slope is a negative velocity.
A long tangent triangle reduces reading error. Markers accept a small range of answers because hand-drawn tangents vary.
A tangent drawn to a position–time graph at $t = 4.0$ s passes through $(2.0\ \text{s},\ 0\ \text{m})$ and $(7.0\ \text{s},\ 60\ \text{m})$. Find the instantaneous velocity at 4.0 s.
A ball is thrown straight up and its height is $s = 20t - 4.9t^2$ (m). Find the average velocity between 1.0 s and 3.0 s.
- With the car at $t = 4$ s and Δt = 2 s, the chord gives an average velocity of 15 m/s. Shrink Δt to 0.1 s: it drops to 12.15 m/s. Turn on the tangent: the instantaneous velocity is 12 m/s.
- Keep shrinking Δt to 0.01 s. How close does the chord get to the tangent?
- Switch to the ball thrown upwards and move $t$ to 2.0 s. The tangent is almost flat. What does that tell you about the ball?
- For the train braking, where is the tangent steepest, and when is it horizontal? Check that the train stops at 10 s.
Test yourself
On a position–time graph, the instantaneous velocity at a point is given by
The tangent's gradient is the rate of change of position at that instant.
A car's position is $s = 1.5t^2$ (m). Its average velocity from $t = 2.0$ s to $t = 6.0$ s is
$s(6) = 54$ m, $s(2) = 6$ m, so $v_{\text{avg}} = 48 \div 4.0 = 12$ m s⁻¹.
For which motion are the average and instantaneous velocities always equal?
Only constant velocity (straight line, constant speed) gives a straight position–time graph.
Point-to-point cameras 30 km apart record a car at 9:00:00 and 9:16:00. The car's average speed is closest to
$30 \div (16/60) = 112.5$ km/h.
Label the chord and the tangent.
Order these steps for finding instantaneous velocity from a graph.
The gradient of a chord on a position–time graph gives the velocity. The gradient of a gives the instantaneous velocity. As the time interval gets smaller, the average velocity gets closer to the velocity. A horizontal tangent means the object is momentarily at .
The position of a cyclist is given by $s = 2.0t^2$ (m) for the first 5.0 s. (a) Calculate the average velocity over the first 4.0 s. (b) Calculate the average velocities from 3.0 s to 3.5 s and from 3.0 s to 3.1 s. (c) Use your answers to estimate the instantaneous velocity at 3.0 s and explain your reasoning.
(a) $s(4.0) = 2.0 \times 16 = 32$ m, so $v_{\text{avg}} = 32 \div 4.0 = 8.0$ m s⁻¹.
(b) $s(3.0) = 18$ m, $s(3.5) = 24.5$ m, $s(3.1) = 19.22$ m. From 3.0 to 3.5 s: $6.5 \div 0.5 = 13$ m s⁻¹. From 3.0 to 3.1 s: $1.22 \div 0.1 = 12.2$ m s⁻¹.
(c) About 12 m s⁻¹. Each shorter interval gives a chord closer to the tangent at 3.0 s, so the average velocity approaches the instantaneous velocity: 13, then 12.2, heading towards 12 m s⁻¹.