Displacement Formula Explained: Definition, Equations, and Examples

Displacement is one of the first big ideas in physics, and it trips up a surprising number of students because it sounds like a fancy word for distance. It is not. Displacement measures where you ended up relative to where you started, in a straight line, with a direction attached. Distance measures how far you actually traveled along the way. Get that distinction right and most of kinematics suddenly makes sense.

In this guide we will define displacement precisely, show every common formula, walk through worked examples step by step, and flag the mistakes that cost easy marks. If you also need related quantities, our velocity formula guide and acceleration formula guide pair naturally with this one.

What is the displacement formula

Displacement is the change in position of an object. The core definition uses the Greek letter delta, which means change:

delta x = x final minus x initial

Here x initial is the starting position and x final is the ending position, both measured along a chosen axis from a fixed reference point called the origin. Because position has a direction, displacement is a vector quantity. A positive value means the object ended up on the positive side of the axis, and a negative value means the negative side. The SI unit is the meter, the same unit used for distance.

It helps to keep the standard symbols straight before you start substituting numbers. Every equation of motion is built from the same small set of letters, so once you recognize them the formulas stop looking intimidating.

A quick example makes it concrete. If you start at the 2 meter mark and walk to the 9 meter mark, your displacement is 9 minus 2, which equals 7 meters in the positive direction. If you then walk back to the 4 meter mark, your displacement for that second leg is 4 minus 9, which equals negative 5 meters, meaning 5 meters in the negative direction.

Displacement vs distance

This is the single most important comparison in the topic. Distance is a scalar, so it only has size and is always positive or zero. Displacement is a vector, so it has size and direction and can be positive, negative, or zero. They are equal only when motion happens in one straight line without reversing.

The round trip row is the classic trap. Run one full lap of a 400 meter track and your distance is 400 meters, but your displacement is zero because you finished exactly where you started. Examiners love this scenario precisely because it separates students who memorized a word from students who understand the concept.

The three equations of motion

When acceleration is constant, you do not always know the start and end positions directly. Instead you might know the initial velocity, final velocity, acceleration, or time, and you want to find the displacement. The three equations of motion, often called the SUVAT equations, let you do exactly that.

In these formulas, s is displacement, u is initial velocity, v is final velocity, a is acceleration, and t is time elapsed.

• s = u t + one half a t squared when you know initial velocity, acceleration, and time.
• s = ((u + v) / 2) t when you know initial velocity, final velocity, and time. This uses the average velocity.
• v squared = u squared + 2 a s when you do not know the time but want to link velocity, acceleration, and displacement.

Each equation leaves out one variable, so you choose the one that uses the three values you already have. The middle formula is just the average of the start and end velocities multiplied by time, which is intuitive once you see it. To work backward from displacement to speed, our velocity calculator and acceleration calculator handle the arithmetic for you.

Choosing the right equation

List what you know and what you want, then pick the formula that contains all four of those symbols. The table below shows which variable each displacement equation leaves out, so you can match it to the value your problem does not give you.

The fourth row is a handy variant of the first that some courses include. It solves for displacement when you know the final velocity instead of the initial one, which happens often in deceleration problems.

Worked example: object under constant acceleration

A car starts from rest and accelerates at 3 meters per second squared for 5 seconds. How far does it travel, and how fast is it going at the end? Because it never reverses, displacement and distance are equal here.

1. Write down the known values. Initial velocity u = 0 (starts from rest), acceleration a = 3 meters per second squared, and time t = 5 seconds.
2. Choose the displacement formula that uses u, a, and t, which is s = u t + one half a t squared.
3. Substitute the numbers. s = (0)(5) + one half times 3 times (5 squared).
4. Simplify the squared term. 5 squared is 25, so the second part is one half times 3 times 25, which is 37.5.
5. Add the parts. s = 0 + 37.5, so the displacement s = 37.5 meters.
6. Find the final velocity with v = u + a t. That is 0 + 3 times 5, which equals 15 meters per second.
7. State the answer with direction. The car is displaced 37.5 meters in the direction of motion and is traveling at 15 meters per second.

As a check, use s = ((u + v) / 2) t. The average of 0 and 15 is 7.5, and 7.5 times 5 equals 37.5 meters. Two independent formulas agree, which confirms the result.

Finding displacement from a velocity time graph

You do not always need a formula. On a velocity time graph, displacement is the area between the line and the time axis. This works for any motion, even when acceleration is not constant, which makes it the most general method of all.

• For motion at constant velocity, the area is a rectangle, so displacement equals velocity times time.
• For constant acceleration starting from rest, the area is a triangle, so displacement equals one half base times height, which is one half times t times v.
• For a trapezium shape, split it into a rectangle plus a triangle, or use the trapezium area rule, and add the pieces together.
• Area below the time axis counts as negative displacement, because the object is moving in the opposite direction.

This is why s = ((u + v) / 2) t works. That formula is simply the area of the trapezium under a straight velocity line, with u and v as the two parallel sides and t as the width. Seeing the geometry behind the algebra makes the equation easy to remember.

Displacement in two dimensions

When an object moves across a plane rather than along a single line, displacement has two components. You find the change in x and the change in y separately, then combine them into a single resultant vector. The magnitude comes from the Pythagorean theorem and the direction comes from the angle of that vector.

For a move from the origin to a point 3 meters east and 4 meters north, the horizontal change is 3 and the vertical change is 4. The magnitude of the displacement is the square root of 3 squared plus 4 squared, which is the square root of 25, or 5 meters. This is exactly the calculation behind the distance formula, which returns the straight line magnitude of a two dimensional displacement.

Angular displacement

For objects that rotate or move along a circular path, the matching idea is angular displacement, the angle swept out around a fixed center. It is written with the Greek letter theta and is measured in radians in SI, where 2 pi radians equals one full turn of 360 degrees.

The link between linear and angular displacement is s = r theta, where s is the arc length traveled, r is the radius, and theta is the angular displacement in radians. So a point on a wheel of radius 2 meters that traces an arc of 10 meters has an angular displacement of 10 divided by 2, which is 5 radians. Counterclockwise is taken as positive and clockwise as negative, mirroring the sign convention for straight line displacement.

Common mistakes to avoid

• Treating displacement as distance. Always ask whether the object reversed direction. If it did, the path length and the net displacement are different numbers.
• Dropping the direction. A displacement answer is incomplete without a sign or a stated direction such as east or upward. Magnitude alone describes distance, not displacement.
• Forgetting to square the time. In s = u t + one half a t squared only the t in the second term is squared. Squaring the whole expression is a frequent error.
• Mixing units. Convert everything to meters and seconds before substituting, or your answer will be off by powers of ten. See our meters in a mile guide when distances are given in miles.
• Using a constant acceleration formula when acceleration changes. The three equations of motion only hold when acceleration is constant. Variable acceleration needs calculus or a graph.

Good to know

In more than one dimension, displacement is found by combining the change in each coordinate as a vector. For two dimensions you compute the change in x and the change in y separately, then find the straight line resultant using the Pythagorean theorem. This is exactly the idea behind the distance formula, which gives the magnitude of a two dimensional displacement between two points.

Average velocity is defined as displacement divided by time, not distance divided by time, which is why a runner who finishes a lap has an average velocity of zero even though their average speed was high. Keeping vectors and scalars in separate mental boxes is the habit that makes the rest of mechanics click.