Average and Instantaneous Velocity in HSC Physics

A car travels 20 m in 4.0 s. Its average velocity is easy to calculate. But what was its velocity exactly 2.0 s after it started?

That question is harder than it looks. Velocity is defined using displacement divided by time, but an instant has no time interval. If the time interval is zero, the calculation seems to become \(0/0\), which tells us nothing.

The solution is one of the most important ideas in kinematics: measure velocity over a short interval, make that interval shorter and shorter, and see what value the measurement approaches. On a displacement-time graph, the same idea appears geometrically as the gradient of a tangent.

Average velocity, instantaneous velocity, graph gradients, and practical measurements are therefore not four separate topics. They are different ways of answering the same question: how quickly is displacement changing with time?

01Start with average velocity

Suppose a student walks 12 m east in 6.0 s.

Before calculating anything, predict the average velocity.

You probably expect \(2.0\text{ m s}^{-1}\) east. That is correct:

\[
v_{\text{av}}=\frac{\Delta x}{\Delta t}
=\frac{12}{6.0}
=2.0\text{ m s}^{-1}\text{ east}
\]

Here:

• \(v_{\text{av}}\) is average velocity in metres per second, \(\text{m s}^{-1}\)
• \(\Delta x\) is displacement in metres, m
• \(\Delta t\) is the time interval in seconds, s
• \(\Delta\) means “change in”

The word average matters. The calculation tells us how quickly displacement changed over the whole interval. It does not tell us whether the student actually moved at \(2.0\text{ m s}^{-1}\) at every moment.

They might have started slowly, sped up, stopped to tie a shoelace, and then sprinted the last few metres.

If displacement and velocity still feel interchangeable, revise distance and displacement first. Velocity depends on displacement, not total distance travelled.

Check your understanding

A runner moves from \(x=-3\text{ m}\) to \(x=9\text{ m}\) in 4.0 s. What is the average velocity?

02Why average velocity is sometimes not enough

Imagine watching a car’s dashboard.

At one moment the speedometer reads 35 km/h. A second later it reads 42 km/h. Later it reads 57 km/h.

The car clearly does not have one velocity for the whole journey. Its velocity is changing.

So suppose we want the velocity at exactly \(t=3.0\text{ s}\).

One possible approach is to measure the car’s displacement between \(2.0\text{ s}\) and \(4.0\text{ s}\), then calculate the average velocity over those two seconds.

That gives us an estimate of the velocity near \(3.0\text{ s}\).

Could we improve it?

Yes. Measure over a smaller interval.

Instead of \(2.0\text{ s}\) to \(4.0\text{ s}\), use \(2.9\text{ s}\) to \(3.1\text{ s}\).

Then perhaps \(2.99\text{ s}\) to \(3.01\text{ s}\).

As the time interval becomes smaller, the average velocity describes the motion increasingly close to the instant we care about.

The value it approaches is the instantaneous velocity.

Formally,

\[
v=\lim_{\Delta t\to 0}\frac{\Delta x}{\Delta t}
\]

You do not normally need to perform limits formally in HSC Physics. The important physical idea is simpler:

Instantaneous velocity is the value approached by average velocity as the measurement interval becomes extremely small.

It is not found by literally setting \(\Delta t=0\). We examine what happens as \(\Delta t\) approaches zero.

03The graph version of the same idea

A displacement-time graph places displacement \(x\) on the vertical axis and time \(t\) on the horizontal axis.

Suppose we choose two points on the graph:

\[
(t_1,x_1)
\]

and

\[
(t_2,x_2)
\]

The gradient between them is

\[
\text{gradient}=\frac{x_2-x_1}{t_2-t_1}
\]

But that is exactly

\[
\frac{\Delta x}{\Delta t}
\]

so the gradient of the line joining two points on a displacement-time graph is the average velocity between those times.

That line is called a secant.

You do not need to become emotionally attached to the word. A secant is simply a straight line joining two points on a curve.

04From a secant to a tangent

Suppose the displacement-time graph is curved because the object is accelerating.

Pick a point at \(t=4.0\text{ s}\). We want the velocity at that instant.

First, join the point at \(t=4.0\text{ s}\) to a point at \(t=6.0\text{ s}\). Calculate the gradient.

Now move the second point closer, perhaps to \(t=5.0\text{ s}\).

Then \(4.5\text{ s}\).

Then \(4.1\text{ s}\).

The secant rotates towards a particular line.

In the limit, it becomes the tangent to the curve at \(t=4.0\text{ s}\).

Therefore:

The gradient of a tangent to a displacement-time graph gives the instantaneous velocity.

This is the graphical version of shrinking the measurement interval towards zero.

A useful mental picture

Imagine taking a photograph of a curved road from high above.

From far away, the road obviously curves. But if you zoom in on one tiny section, that small piece begins to look almost straight.

Instantaneous velocity works similarly. Over a large time interval, the motion may change substantially. Over a sufficiently tiny interval, the motion near one instant behaves almost like constant-velocity motion.

The analogy has a limit. The actual displacement-time curve does not literally become straight. We are examining its local gradient.

Check your understanding

A displacement-time graph is getting steeper as time increases. What can you say about the object’s velocity?

06Estimating instantaneous velocity using a short interval

Now use the same motion,

\[
x=2t^2
\]

but estimate the instantaneous velocity at \(t=3.0\text{ s}\).

We could measure the average velocity between \(2.9\text{ s}\) and \(3.1\text{ s}\).

07What a tangent calculation actually looks like

Students are often told to “draw a tangent” and then jump straight to a gradient calculation. It helps to slow that process down.

Suppose you have a curved displacement-time graph and need the instantaneous velocity at \(t=6\text{ s}\).

First, draw a straight line that touches the curve at the point corresponding to \(6\text{ s}\) and follows the direction of the curve at that point.

Then choose two convenient points on the tangent line.

They do not have to be points on the original data curve.

For example, suppose the tangent passes through:

\[
(2\text{ s},4\text{ m})
\]

and

\[
(10\text{ s},28\text{ m})
\]

Then

\[
v=\text{gradient}
=\frac{28-4}{10-2}
=\frac{24}{8}
=3.0\text{ m s}^{-1}
\]

The instantaneous velocity at \(t=6\text{ s}\) is therefore approximately \(3.0\text{ m s}^{-1}\).

Why should you use points far apart on the tangent?

You might think the best method is to use two points extremely close to the point of tangency.

Usually, that makes the measurement worse.

Suppose your ruler position is uncertain by about \(1\text{ mm}\). If your two chosen points are only a few millimetres apart, that small uncertainty becomes a large fraction of the measured rise and run.

Using two well-separated points on the same tangent reduces the percentage effect of your reading uncertainty.

This does not mean using points far apart on the original curve. That would give an average velocity. The points must lie on the straight tangent you have drawn.

Check your understanding

A tangent to a displacement-time graph passes through \((1.0\text{ s},2.0\text{ m})\) and \((7.0\text{ s},20.0\text{ m})\). Estimate the instantaneous velocity at the point where the tangent touches the curve.

08How do you measure instantaneous velocity in a real experiment?

Here is the experimental problem.

Suppose a trolley is rolling down a ramp. You want its velocity at one particular point.

You cannot place a stopwatch at one instant and measure “zero seconds” of motion. Real equipment always needs a finite distance or finite time interval.

So practical experiments approximate instantaneous velocity by making that interval small.

09Method 1: A light gate and interrupt card

A common setup uses a light gate.

Picture the apparatus from left to right.

A trolley moves along a track. Attached to the trolley is a rectangular card of known length. Further along the track is a light gate containing a light source and detector.

As the trolley moves through the gate:

1. the front edge of the card blocks the light beam
2. the timer starts
3. the card continues moving through the gate
4. the rear edge clears the beam
5. the timer stops

Suppose the card has length \(L\), and blocks the light for time \(\Delta t\).

The measured velocity is

\[
v\approx\frac{L}{\Delta t}
\]

Why is this only an approximation to instantaneous velocity?

Because the trolley moves through a small but non-zero distance \(L\). The calculation therefore gives the average velocity while the card passes through the gate.

If \(L\) is short enough, and the trolley’s velocity does not change much while the card is passing, this average is a good approximation to the velocity at the gate.

10Why not make the card infinitely short?

If a shorter card gives a more local measurement, it might seem that we should make the card as short as physically possible.

There is a trade-off.

A very short card reduces the time interval, which can make the approximation to instantaneous velocity better. But it also makes timing uncertainty more important.

Suppose the timer uncertainty is \(0.001\text{ s}\).

If the card blocks the gate for \(0.100\text{ s}\), the timing uncertainty is about 1% of the measured interval.

If it blocks the gate for only \(0.005\text{ s}\), that same \(0.001\text{ s}\) uncertainty is 20% of the measurement.

So practical experimental design balances two competing goals:

• make the interval short enough that velocity changes very little
• keep the interval large enough to measure accurately

This is a general experimental idea, not just a velocity trick.

11Method 2: Ticker timer measurements

Before electronic motion sensors became common, ticker timers offered another way to estimate velocity.

A narrow paper tape runs through a ticker timer. The tape is attached to a moving trolley. As the trolley pulls the tape through the device, the timer places dots on the tape at equal time intervals.

If the timer operates at \(50\text{ Hz}\), it makes 50 dots each second.

The time between successive dots is therefore

\[
\Delta t=\frac{1}{50}=0.020\text{ s}
\]

Now look at the dot spacing.

Closely spaced dots mean the trolley moved only a small distance during each \(0.020\text{ s}\) interval, so its speed was relatively low.

Widely spaced dots mean greater displacement in the same time, so the speed was higher.

If the gaps get progressively larger, the trolley is accelerating.

A historical measurement problem

The difficulty is that a single gap still gives average velocity over \(0.020\text{ s}\), not velocity at a mathematical instant.

One useful improvement is to measure across several adjacent intervals centred on the point of interest.

For example, if a particular dot represents the time we care about, we can measure from a dot just before it to a dot just after it.

This produces an average over a small interval centred approximately on the desired instant.

The method does not create a truly zero-time measurement. Instead, it uses the same idea we developed earlier: a small average can approximate an instantaneous value.

12Modern motion sensors do the same thing more quickly

A motion sensor may appear to give instantaneous position and velocity directly.

It doesn’t escape the underlying measurement problem.

The sensor records position at a series of discrete times. Software can then estimate velocity by comparing nearby position measurements.

For example, suppose a sensor records:

To estimate velocity at \(2.00\text{ s}\), use the measurements on either side:

\[
v\approx\frac{3.54-3.42}{2.02-1.98}
=\frac{0.12}{0.04}
=3.0\text{ m s}^{-1}
\]

Again, this is a short-interval estimate centred on the time of interest.

A graphing program might perform this automatically, but the physics underneath has not changed.

13The mistake students make with displacement-time graphs

Here is a tempting prediction:

“If the displacement is large, the velocity must be large.”

It sounds reasonable. An object far from the origin feels like it must have travelled quickly.

But displacement is represented by the height of a displacement-time graph. Velocity is represented by its gradient.

Those are different properties.

Consider a horizontal line at \(x=100\text{ m}\).

The object has a displacement of \(100\text{ m}\), but

\[
\text{gradient}=0
\]

so

\[
v=0
\]

The object is 100 m from the origin and completely stationary.

Now consider a graph passing through \(x=0\text{ m}\) with a very steep positive gradient.

At that instant, the displacement is zero but the velocity could be large.

Position tells you where the object is.

Gradient tells you how its position is changing.

Check your understanding

At \(t=5\text{ s}\), an object’s displacement-time graph crosses the time axis with a steep negative gradient. What does this tell you?

14Another common mistake: speed and velocity are not identical

Suppose a displacement-time graph has a gradient of

\[
-6.0\text{ m s}^{-1}
\]

The instantaneous velocity is

\[
-6.0\text{ m s}^{-1}
\]

but the instantaneous speed is

\[
6.0\text{ m s}^{-1}
\]

Speed is the magnitude of velocity. Velocity includes direction.

That distinction becomes especially important once graphs cross zero or change gradient sign. The full relationship is covered in speed and velocity.

15Reading motion directly from displacement-time graphs

A displacement-time graph can tell you much more than one velocity value.

The last case deserves care.

Suppose a ball moves upwards. Its displacement increases, but its upward velocity decreases. At the highest point, the tangent to the displacement-time graph is horizontal.

So at that instant,

\[
v=0
\]

Immediately afterwards, the graph slopes downwards, so the velocity has become negative.

A zero instantaneous velocity therefore does not mean the object must remain stationary. It may simply be at a turning point.

16Average velocity can hide a lot of motion

Imagine walking 20 m east and then 20 m west, ending where you started.

Suppose the trip takes 40 s.

Your total displacement is

\[
\Delta x=0
\]

so

\[
v_{\text{av}}=\frac{0}{40}=0\text{ m s}^{-1}
\]

Does that mean you were stationary for 40 s?

Obviously not.

Average velocity only compares the starting and finishing positions over the chosen interval. It does not describe everything that happened in between.

This is why the interval you choose matters.

Over the full 40 s, the average velocity is zero.

During the first half, it is positive.

During the second half, it is negative.

At individual instants, the velocity may take many different values.

Check your understanding

A car completes one lap of a circular track and finishes exactly where it started. Is its average velocity zero?

17A harder graph calculation

Suppose a displacement-time graph is described by

\[
x=t^2+2t
\]

where \(x\) is measured in metres and \(t\) in seconds.

Estimate the instantaneous velocity at \(t=4.0\text{ s}\) using the interval from \(3.9\text{ s}\) to \(4.1\text{ s}\).

18What if the tangent slopes downwards?

Nothing special happens mathematically.

Suppose a tangent passes through

\[
(3\text{ s},12\text{ m})
\]

and

\[
(8\text{ s},-3\text{ m})
\]

Then

\[
v=\frac{-3-12}{8-3}
=\frac{-15}{5}
=-3.0\text{ m s}^{-1}
\]

The negative value means motion in the negative direction.

Students sometimes remove the negative sign because they think velocity “can’t be negative”. That turns velocity into speed and loses directional information.

19How average and instantaneous velocity fit together

The easiest way to organise the topic is this:

The same core idea keeps reappearing:

\[
\text{velocity}=\text{rate of change of displacement}
\]

The only question is over what interval?

For average velocity, the interval is finite.

For instantaneous velocity, we consider what the average velocity approaches as that interval becomes extremely small.

20The next step: acceleration

Once velocity can change from one instant to the next, another question appears naturally.

How quickly is velocity itself changing?

That quantity is acceleration:

\[
a_{\text{av}}=\frac{\Delta v}{\Delta t}
\]

The structure is deliberately familiar.

Average velocity describes how displacement changes with time.

Acceleration describes how velocity changes with time.

That is why kinematics becomes much easier once you stop treating every equation as a separate formula to memorise. The graphs, interval calculations, tangents, light gates, ticker timers, and motion sensors are all measuring different versions of the same thing: how one physical quantity changes with another.