Formula of Displacement: Δx = x_f − x

The formula of displacement in its simplest form is $\Delta x = x_f - x_i$ : final position minus initial position. When an object moves with constant acceleration, the same displacement can be found from the second equation of motion, $s = ut + \frac{1}{2}at^2$ , without knowing the final position at all.

Displacement is a vector. It tells you how far, and in which direction, an object ends up from where it began, measured in a straight line. A taxi that drives 6 km around a city and stops 2 km east of its starting rank has a displacement of 2 km east. In Class 9 and Class 11 books it is written s; US books use Δx for motion along a line.

Quick facts

Definition
$\Delta x = x_f - x_i$
Constant acceleration
$s = ut + \frac{1}{2}at^2$
SI unit
metre (m)
Dimensional formula
[M⁰ L¹ T⁰]
Type
Vector: can be positive, negative or zero
From a graph
Signed area under a velocity time graph

What is the formula of displacement? The forms you need

Two ways to get displacement
Form	When to use it	Symbols
$\Delta x = x_f - x_i$	You know the start and end positions	x positions in m
$s = ut + \frac{1}{2}at^2$	Acceleration is constant and you know u, a and t	u in m/s, a in m/s², t in s
$s = \frac{(u + v)}{2}\, t$	Constant acceleration, both velocities known	v final velocity in m/s
$s = \frac{v^2 - u^2}{2a}$	Constant acceleration, time not given	from v² = u² + 2as

In $s = ut + \frac{1}{2}at^2$ , the first term is how far the object would go at its starting velocity, and the second is the extra distance the acceleration adds. If the object starts from rest, u = 0 and the formula shrinks to $s = \frac{1}{2}at^2$ .

Worked examples

Example 1: a walk in two directions

A student walks 4.0 m east, then 3.0 m north. Find the displacement.

The two legs are at right angles, so use Pythagoras: $s = \sqrt{4.0^2 + 3.0^2} = \sqrt{25} = 5.0\ \text{m}$ .

Direction: $\tan\theta = \frac{3.0}{4.0}$ , so $\theta \approx 37^\circ$ north of east.

The distance walked is 7.0 m, but the displacement is 5.0 m at about 37° north of east.

Example 2: a car pulling away

A car starts from rest and accelerates at 3.0 m/s² for 6.0 s. Find its displacement.

$s = 0 + \tfrac{1}{2}(3.0)(6.0)^2 = 1.5 \times 36 = 54\ \text{m}$

Example 3: a ball thrown upward

A ball is thrown straight up at 15 m/s. Find its displacement after 2.0 s. Take up as positive and g = 9.8 m/s².

Here a = −9.8 m/s².

$s = 15(2.0) + \tfrac{1}{2}(-9.8)(2.0)^2 = 30 - 19.6 = 10.4\ \text{m}$

The ball is 10.4 m above the launch point. It has already passed its highest point (about 11.5 m) and is on the way down, so the distance travelled is larger than 10.4 m.

Formula of displacement for Class 9: the graph method

On a velocity time graph, the area between the line and the time axis is the displacement. Area above the axis counts as positive and area below counts as negative. For a straight line starting at u and ending at v after t seconds, the shape is a trapezium with area (u + v)t/2, which is exactly the third row of the table above.

On a position time graph, the displacement between two moments is simply the change in the vertical reading. Both graph ideas are tested in board exams, often with a sketch to label.

Units and dimensions

Displacement is a length, measured in metres, with dimensions of pure length: [M⁰ L¹ T⁰]. Every term in a displacement equation must reduce to a length. Check it on the second equation: ut has (m/s)(s) = m, and at² has (m/s²)(s²) = m. The number ½ has no dimension. The same bookkeeping, done for every quantity, appears on the dimensional formula of displacement page.

Displacement compared with distance

Distance adds up every metre of the path and is never negative. Displacement only compares the end with the start, so it can be zero after a long trip. If the object never turns and never leaves a straight track, both have the same size. Once the path bends or turns back, the size of the displacement is smaller than the distance. See the distance formula physics page for the scalar version.

Displacement divided by time gives velocity, which is why velocity is a vector too.

Traps in displacement questions

Adding legs that point different ways. 4 m east and 3 m north do not make 7 m of displacement. Add them as vectors.
Forgetting the sign of a. For a body thrown upward with up as positive, a = −9.8 m/s².
Using s = ut + ½at² when the acceleration changes. It only holds for constant a.
Reporting a size with no direction. Give the angle or a compass direction.

Need v or u instead of s? The kinematics formulas page lists the partner equations.

Frequently asked questions

What is the formula for displacement in physics?

Displacement = final position − initial position, Δx = x_f − x_i. With constant acceleration you can use s = ut + ½at².

Which SI unit is used for displacement?

The metre (m), the same unit as distance. The difference is that displacement also has a direction.

Can displacement be zero?

Yes. If an object comes back to where it started, its displacement is zero, however far it travelled.

Can displacement be negative?

Yes. A negative displacement means the object ended up on the negative side of its starting point, along the axis you chose.

Is displacement ever larger than distance?

No. The size of the displacement is at most equal to the distance. They are equal only for straight line motion with no turning back.