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Dimensional Formula of Displacement

Dimensional formulas of physical quantitiesClass 9Class 11Updated 1 Nov 2026

[s]=[M0L1T0][s] = [M^{0} L^{1} T^{0}]
SI unit
metre (m)
Dimensions
[M⁰ L¹ T⁰]
where
  • s displacement m
  • L dimension of length m

The dimensional formula of displacement is [M⁰ L¹ T⁰], usually just [L]. Displacement is the straight line change in position of a body from start to finish, so it is a length, and length belongs to the SI base set.

The direction that makes displacement a vector does not show up in the bracket. A car that drives 5 km east and one that drives 5 km west have opposite displacements but identical dimensions. This makes [L] the most useful check in kinematics: each term of a motion equation that gives a position has to come out as [L]. For the formulas themselves, open the displacement formula guide.

Quick facts

  • Dimensional formula

    [M⁰ L¹ T⁰], or [L]

  • Defining relation

    s⃗=r⃗final−r⃗initial\vec{s} = \vec{r}_{final} - \vec{r}_{initial}

  • SI unit

    metre (m)

  • Type

    Vector, with size and direction

  • Common symbols

    s, x, Δx, d

  • Same bracket

    Distance, wavelength, radius, amplitude

Why the dimensional formula of displacement is [L]

  1. Displacement is final position minus initial position: Δx=x2−x1\Delta x = x_2 - x_1.
  2. Each position is measured in metres from an origin, so [x]=[L][x] = [L].
  3. A difference of two lengths is still a length: [Δx]=[L][\Delta x] = [L].
  4. With all powers shown, that is [M⁰ L¹ T⁰].

The same result comes from velocity. Average velocity is displacement over time, v=stv = \frac{s}{t}, so s=vts = vt, and [LT−1][T]=[L][LT^{-1}][T] = [L]. The velocity bracket is built from this one, not the other way round, since length is a base quantity.

Checking the equations of motion

Example: test s=ut+12at2s = ut + \frac{1}{2}at^{2} term by term.

[ut]=[LT−1][T]=[L][ut] = [LT^{-1}][T] = [L].

[at2]=[LT−2][T2]=[L][at^{2}] = [LT^{-2}][T^{2}] = [L]. The 1/2 is just a number.

Every term is [L], so the equation is consistent.

Example: someone proposes s=ut2+ats = ut^{2} + at. Could that be right?

[ut2]=[LT][ut^{2}] = [LT] and [at]=[LT−1][at] = [LT^{-1}]. Neither term is a length, so the equation is wrong.

Be careful in the other direction. s=ut+at2s = ut + at^{2}, without the 1/2, also gives [L] in every term, yet it is wrong. Dimensions catch missing powers, never missing numbers. The full set of motion equations sits on the kinematics formulas hub.

Using [L] to predict a formula

Example: a ball is dropped from rest. Assume the distance fallen depends only on g and the time t. Find the form of the formula.

Write s=Cgatbs = C g^{a} t^{b}, with C a pure number.

Dimensions: [L]=[LT−2]a[T]b=[LaT−2a+b][L] = [LT^{-2}]^{a}[T]^{b} = [L^{a}T^{-2a+b}].

Match powers. L: a = 1. T: −2a + b = 0, so b = 2.

Result: s=Cgt2s = C g t^{2}. Experiment or calculus then shows C = 1/2.

This method of dimensions is a standard Class 11 question type. It finds the powers but never the constant, and it fails when a quantity depends on a sum such as u + at.

Lengths that share the bracket

All of these are measured in metres
QuantityScalar or vectorNote
DisplacementVectorShortest path, start to end
DistanceScalarTotal path length
WavelengthScalarCrest to crest
AmplitudeScalarLargest displacement in an oscillation
Radius of gyrationScalarFrom I = Mk²
Focal lengthSigned scalarSign set by convention

Displacement and distance share [L] but differ in meaning. Walk once round a 400 m track and your distance is 400 m while your displacement is zero. Dimensions cannot tell these apart, so read the question to see which is wanted.

Unit conversions and slips

Since the bracket is pure length, converting displacement only needs length factors: 1 km = 10³ m, 1 cm = 10⁻² m, 1 Å = 10⁻¹⁰ m, and one light year is about 9.46 × 10¹⁵ m.

  • Writing [LT⁻¹]. That is velocity.
  • Expecting a direction in the bracket. Dimensions never carry direction.
  • Calling the light year a time. It is the distance light covers in a year, so its bracket is [L].
  • Thinking zero displacement means no dimension. The value is zero; the bracket stays [L].

The dimensional formulas list starts with the base quantities, so displacement, mass and time sit at the very top of it.

Frequently asked questions

What is the dimensional formula of displacement?

Displacement carries [M⁰ L¹ T⁰], which is just [L]. It is a change in position, measured as a length.

Do displacement and distance have the same dimensional formula?

They do; each reduces to [L] and is measured in metres. Displacement is a vector and distance is a scalar, a difference that dimensions do not show.

Is displacement a vector?

It is. Displacement carries a magnitude in metres plus a direction from start point to end point.

Which unit should a displacement answer carry?

The metre, m. Turn km, cm or mm into metres before you put them into any SI formula.

Is a negative displacement allowed?

Certainly. The minus just says the final position lies behind the starting point on whatever axis you called positive. The bracket is still [L].