Average velocity, average speed and instantaneous velocity

How fast, and which way?

A car's speedometer reads $50\ \text{km/h}$. A bus timetable says the bus averages $30\ \text{km/h}$ between two towns. A cricket commentator says the ball left the bat at $120\ \text{km/h}$. Three different ideas hide behind the word "speed". This lesson sorts them out.

Average speed and average velocity

Over any stretch of time $\Delta t$:

Average speed is a scalar, never negative. Average velocity has a sign: plus means the body went, overall, in the positive direction.

A worked example. A cyclist rides $3\ \text{km}$ east in $10$ minutes, then $1\ \text{km}$ west in $5$ minutes. The whole time is $15\ \text{min} = 0.25\ \text{h}$.

1. Average speed. $\dfrac{3\ \text{km} + 1\ \text{km}}{0.25\ \text{h}} = 16\ \text{km/h}$.
2. Average velocity. $\dfrac{3\ \text{km} - 1\ \text{km}}{0.25\ \text{h}} = 8\ \text{km/h}$, east.

The two agree only when the body never turns back.

Watch out: averaging two speeds. A bus goes $30\ \text{km}$ at $50\ \text{km/h}$ and the next $30\ \text{km}$ at $75\ \text{km/h}$. Its average speed is not $62.5\ \text{km/h}$. The times are $0.6\ \text{h}$ and $0.4\ \text{h}$, so the average is $\dfrac{60}{1.0} = 60\ \text{km/h}$. The bus spends longer at the slower speed, so the average is pulled towards it. Always go back to the definition: whole distance over whole time.

The velocity at an instant

A speedometer does not show an average over the trip. It shows how fast the car goes right now. How can we give a meaning to "right now", when a speed always needs a distance and a time?

The idea: take the average velocity over a very short time after the instant, and let that time shrink towards zero.

On a position-time graph, the average velocity from P to Q is the slope of the straight line through P and Q. As Q slides towards P, that line turns into the tangent at P.

So: the velocity at an instant is the slope of the tangent to the position-time graph at that instant. Its size is the speed at that instant, the number on a speedometer.

With calculus

When the position is given as a formula, differentiate it. Two rules do most of the work here: the rate of change of $t^n$ is $n\,t^{n-1}$, and of a constant is $0$.

A worked example. A particle moves with $x = 3t^2 - 2t$ (in metres, $t$ in seconds). Find its velocity at $t = 4\ \text{s}$.

1. Differentiate. $v = \dfrac{dx}{dt} = 6t - 2$.
2. Put in the time. $v = 24\ \text{m/s} - 2\ \text{m/s} = 22\ \text{m/s}$.

Think first. Is $22\ \text{m/s}$ the same as $\dfrac{x}{t} = \dfrac{40\ \text{m}}{4\ \text{s}} = 10\ \text{m/s}$? No: $\dfrac{x}{t}$ is the average velocity over the first $4\ \text{s}$. The particle speeds up, so it is moving much faster at the end than on average.

Practise with new numbers

Try it in Python

The program takes $x = t^2$ and works out the average velocity after $t = 3\ \text{s}$ over shorter and shorter intervals.

Remember these

• Average speed $= \dfrac{\text{path length}}{\text{time}}$; average velocity $= \dfrac{\Delta x}{\Delta t}$, with a sign.
• Never average two speeds: divide the whole distance by the whole time.
• Velocity at an instant: $v = \dfrac{dx}{dt}$, the slope of the tangent. Speed is its size.

Check yourself

1. A runner completes one lap of a $400\ \text{m}$ track in $80\ \text{s}$. Find her average speed and her average velocity.

2. A car goes the first $15\ \text{km}$ of a trip at $30\ \text{km/h}$ and the next $15\ \text{km}$ at $90\ \text{km/h}$. Find its average speed.

3. For $x = 5t^2$ (SI units), find the velocity at $t = 2\ \text{s}$, and the average velocity over the first $2\ \text{s}$.

4. What does a car's speedometer show: the average speed, the speed at that instant, or the velocity?