Average Velocity Formula: Δx/Δt, (u + v)/2 and Examples

The average velocity formula is $v_{avg} = \frac{\Delta x}{\Delta t}$ : the total displacement of a trip divided by the total time the trip took. Only where you started and where you finished matter. Every stop, turn and burst of speed along the way is ignored.

Class 9 books write it as average velocity = total displacement / total time. In Class 11, and in US courses, the same rule appears with $\Delta x = x_f - x_i$ . For the special case of constant acceleration there is a shortcut, $v_{avg} = \frac{u + v}{2}$ , and knowing when that shortcut is allowed is the most tested idea on this page. It sits with the other kinematics formulas and grows out of the basic velocity formula.

Quick facts

Formula
$v_{avg} = \frac{\Delta x}{\Delta t}$
Constant acceleration only
$v_{avg} = \frac{u + v}{2}$
SI unit
m/s
Dimensional formula
[M⁰ L¹ T⁻¹]
Type
Vector, in the direction of the net displacement
Round trip
Zero, because the displacement is zero

Average velocity formula physics: the symbols

What each term in v_avg = Δx / Δt stands for
Term	Meaning	SI unit
$v_{avg}$	Average velocity over the whole interval	m/s
$\Delta x = x_f - x_i$	Net displacement: final position minus starting position	m
$\Delta t = t_f - t_i$	Total time, including any time spent stopped	s

Two details trip students up. First, $\Delta x$ can be negative, and then the average velocity is negative too. Second, $\Delta t$ includes rests. If a bus waits 2 minutes at a stop, those 2 minutes belong in the denominator.

Solved examples

Example 1: a bus on a straight road

A bus moves 4.8 km east in 10 minutes. Find its average velocity in m/s.

Convert: Δx = 4800 m, Δt = 10 × 60 = 600 s.

$v_{avg} = \frac{4800}{600} = 8.0\ \text{m/s east}$

Example 2: a trip with a turn back

A cyclist rides 3.0 km north in 600 s, then 1.2 km south in 300 s. Find the average velocity and the average speed.

Net displacement = 3.0 − 1.2 = 1.8 km north = 1800 m. Total time = 900 s. Total distance = 4.2 km = 4200 m.

$v_{avg} = \frac{1800}{900} = 2.0\ \text{m/s north}$

$\text{average speed} = \frac{4200}{900} \approx 4.7\ \text{m/s}$

The two answers differ because the cyclist doubled back.

Example 3: uniform acceleration

A car speeds up evenly from 10 m/s to 30 m/s in 8.0 s. Find its average velocity and its displacement.

Acceleration is constant, so $v_{avg} = \frac{10 + 30}{2} = 20\ \text{m/s}$ .

Displacement $= v_{avg} \times t = 20 \times 8.0 = 160\ \text{m}$ .

Check with $s = ut + \frac{1}{2}at^2$ : a = 2.5 m/s², so s = 80 + 80 = 160 m.

When the (u + v)/2 shortcut works

The mean of the starting and final velocities equals the average velocity only when acceleration stays constant. On a velocity time graph that motion is a straight line, and the average height of a straight line is halfway between its ends.

If acceleration changes, the shortcut fails. Here is the classic trap. A van drives 15 km at 20 km/h and then a further 15 km, in the same direction, at 60 km/h. The first leg takes 0.75 h and the second 0.25 h, so the whole trip lasts 1.0 h.

$v_{avg} = \frac{30\ \text{km}}{1.0\ \text{h}} = 30\ \text{km/h}$

The result is 30 km/h rather than the tempting 40 km/h, because the van spends three times as long at the slower speed. For equal distances the correct result is the harmonic mean, 2v₁v₂ / (v₁ + v₂). For equal times the simple mean is correct.

Before you halve u + v, ask: is the acceleration constant? If you are not told so, use total displacement over total time.

Average versus instantaneous velocity

Average velocity describes a whole interval. Instantaneous velocity describes one moment, the reading a speedometer with a direction would give. On a position time graph, the average velocity between two points is the slope of the straight chord joining them. The instantaneous value is the gradient of the line that just touches the curve at one point.

As the interval shrinks, the chord turns into the tangent and the average becomes the instantaneous value. That limit is how calculus based courses define instantaneous velocity, v = dx/dt.

Average velocity, average speed and average acceleration

Three averages that look alike
Quantity	Formula	Scalar or vector
Average velocity	$\frac{\Delta x}{\Delta t}$	Vector
Average speed	$\frac{\text{total distance}}{\text{total time}}$	Scalar
Average acceleration	$\frac{\Delta v}{\Delta t}$	Vector

The size of the average velocity is never larger than the average speed. They are equal only when the object moves in one direction along a straight line without reversing.

Errors that lose marks

Using distance instead of displacement. For a round trip the average velocity is zero, whatever the speed was.
Dropping rest time. Every second between start and finish counts in Δt.
Averaging speeds blindly. The simple mean works for equal times or constant acceleration, not for equal distances.
No direction in the answer. Write north, east, or a sign. A size alone is a speed.

Frequently asked questions

What is the formula of average velocity?

Average velocity = total displacement / total time, written v_avg = Δx / Δt. If the acceleration is constant you can also use (u + v) / 2.

What is the average velocity formula for Class 9?

Class 9 uses average velocity = (initial velocity + final velocity) / 2 for uniformly changing velocity, and total displacement / total time in general. Use the second one unless the question says the acceleration is uniform.

Can average velocity be zero when speed is not?

Yes. If an object returns to its starting point, its displacement is zero, so the average velocity is zero. Its average speed is still positive because it covered a distance.

What is the difference between average velocity and average speed?

Average velocity uses displacement and has a direction. Average speed uses total path length and has no direction. On a path that turns back, average speed is the larger of the two.

Can average velocity be negative?

Yes. A negative sign simply says the net displacement runs the other way from the axis you picked as positive.