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Physics library

Course: physics library   >   unit 1.

  • Intro to vectors and scalars
  • Introduction to reference frames
  • What is displacement?
  • Calculating average velocity or speed
  • Solving for time

Displacement from time and velocity example

  • Instantaneous speed and velocity
  • What is velocity?
  • Position vs. time graphs
  • What are position vs. time graphs?
  • Average velocity and average speed from graphs
  • Instantaneous velocity and instantaneous speed from graphs

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Video transcript

  • 3.1 Position, Displacement, and Average Velocity
  • Introduction
  • 1.1 The Scope and Scale of Physics
  • 1.2 Units and Standards
  • 1.3 Unit Conversion
  • 1.4 Dimensional Analysis
  • 1.5 Estimates and Fermi Calculations
  • 1.6 Significant Figures
  • 1.7 Solving Problems in Physics
  • Key Equations
  • Conceptual Questions
  • Additional Problems
  • Challenge Problems
  • 2.1 Scalars and Vectors
  • 2.2 Coordinate Systems and Components of a Vector
  • 2.3 Algebra of Vectors
  • 2.4 Products of Vectors
  • 3.2 Instantaneous Velocity and Speed
  • 3.3 Average and Instantaneous Acceleration
  • 3.4 Motion with Constant Acceleration
  • 3.5 Free Fall
  • 3.6 Finding Velocity and Displacement from Acceleration
  • 4.1 Displacement and Velocity Vectors
  • 4.2 Acceleration Vector
  • 4.3 Projectile Motion
  • 4.4 Uniform Circular Motion
  • 4.5 Relative Motion in One and Two Dimensions
  • 5.2 Newton's First Law
  • 5.3 Newton's Second Law
  • 5.4 Mass and Weight
  • 5.5 Newton’s Third Law
  • 5.6 Common Forces
  • 5.7 Drawing Free-Body Diagrams
  • 6.1 Solving Problems with Newton’s Laws
  • 6.2 Friction
  • 6.3 Centripetal Force
  • 6.4 Drag Force and Terminal Speed
  • 7.2 Kinetic Energy
  • 7.3 Work-Energy Theorem
  • 8.1 Potential Energy of a System
  • 8.2 Conservative and Non-Conservative Forces
  • 8.3 Conservation of Energy
  • 8.4 Potential Energy Diagrams and Stability
  • 8.5 Sources of Energy
  • 9.1 Linear Momentum
  • 9.2 Impulse and Collisions
  • 9.3 Conservation of Linear Momentum
  • 9.4 Types of Collisions
  • 9.5 Collisions in Multiple Dimensions
  • 9.6 Center of Mass
  • 9.7 Rocket Propulsion
  • 10.1 Rotational Variables
  • 10.2 Rotation with Constant Angular Acceleration
  • 10.3 Relating Angular and Translational Quantities
  • 10.4 Moment of Inertia and Rotational Kinetic Energy
  • 10.5 Calculating Moments of Inertia
  • 10.6 Torque
  • 10.7 Newton’s Second Law for Rotation
  • 10.8 Work and Power for Rotational Motion
  • 11.1 Rolling Motion
  • 11.2 Angular Momentum
  • 11.3 Conservation of Angular Momentum
  • 11.4 Precession of a Gyroscope
  • 12.1 Conditions for Static Equilibrium
  • 12.2 Examples of Static Equilibrium
  • 12.3 Stress, Strain, and Elastic Modulus
  • 12.4 Elasticity and Plasticity
  • 13.1 Newton's Law of Universal Gravitation
  • 13.2 Gravitation Near Earth's Surface
  • 13.3 Gravitational Potential Energy and Total Energy
  • 13.4 Satellite Orbits and Energy
  • 13.5 Kepler's Laws of Planetary Motion
  • 13.6 Tidal Forces
  • 13.7 Einstein's Theory of Gravity
  • 14.1 Fluids, Density, and Pressure
  • 14.2 Measuring Pressure
  • 14.3 Pascal's Principle and Hydraulics
  • 14.4 Archimedes’ Principle and Buoyancy
  • 14.5 Fluid Dynamics
  • 14.6 Bernoulli’s Equation
  • 14.7 Viscosity and Turbulence
  • 15.1 Simple Harmonic Motion
  • 15.2 Energy in Simple Harmonic Motion
  • 15.3 Comparing Simple Harmonic Motion and Circular Motion
  • 15.4 Pendulums
  • 15.5 Damped Oscillations
  • 15.6 Forced Oscillations
  • 16.1 Traveling Waves
  • 16.2 Mathematics of Waves
  • 16.3 Wave Speed on a Stretched String
  • 16.4 Energy and Power of a Wave
  • 16.5 Interference of Waves
  • 16.6 Standing Waves and Resonance
  • 17.1 Sound Waves
  • 17.2 Speed of Sound
  • 17.3 Sound Intensity
  • 17.4 Normal Modes of a Standing Sound Wave
  • 17.5 Sources of Musical Sound
  • 17.7 The Doppler Effect
  • 17.8 Shock Waves
  • B | Conversion Factors
  • C | Fundamental Constants
  • D | Astronomical Data
  • E | Mathematical Formulas
  • F | Chemistry
  • G | The Greek Alphabet

Learning Objectives

By the end of this section, you will be able to:

  • Define position, displacement, and distance traveled.
  • Calculate the total displacement given the position as a function of time.
  • Determine the total distance traveled.
  • Calculate the average velocity given the displacement and elapsed time.

When you’re in motion, the basic questions to ask are: Where are you? Where are you going? How fast are you getting there? The answers to these questions require that you specify your position, your displacement, and your average velocity—the terms we define in this section.

To describe the motion of an object, you must first be able to describe its position ( x ): where it is at any particular time . More precisely, we need to specify its position relative to a convenient frame of reference. A frame of reference is an arbitrary set of axes from which the position and motion of an object are described. Earth is often used as a frame of reference, and we often describe the position of an object as it relates to stationary objects on Earth. For example, a rocket launch could be described in terms of the position of the rocket with respect to Earth as a whole, whereas a cyclist’s position could be described in terms of where she is in relation to the buildings she passes Figure 3.2 . In other cases, we use reference frames that are not stationary but are in motion relative to Earth. To describe the position of a person in an airplane, for example, we use the airplane, not Earth, as the reference frame. To describe the position of an object undergoing one-dimensional motion, we often use the variable x . Later in the chapter, during the discussion of free fall, we use the variable y .


If an object moves relative to a frame of reference—for example, if a professor moves to the right relative to a whiteboard Figure 3.3 —then the object’s position changes. This change in position is called displacement . The word displacement implies that an object has moved, or has been displaced. Although position is the numerical value of x along a straight line where an object might be located, displacement gives the change in position along this line. Since displacement indicates direction, it is a vector and can be either positive or negative, depending on the choice of positive direction. Also, an analysis of motion can have many displacements embedded in it. If right is positive and an object moves 2 m to the right, then 4 m to the left, the individual displacements are 2 m and −4 −4 m, respectively.

Displacement Δ x Δ x is the change in position of an object:

where Δ x Δ x is displacement, x f x f is the final position, and x 0 x 0 is the initial position.

We use the uppercase Greek letter delta (Δ) to mean “change in” whatever quantity follows it; thus, Δ x Δ x means change in position (final position less initial position). We always solve for displacement by subtracting initial position x 0 x 0 from final position x f x f . Note that the SI unit for displacement is the meter, but sometimes we use kilometers or other units of length. Keep in mind that when units other than meters are used in a problem, you may need to convert them to meters to complete the calculation (see Appendix B ).

Objects in motion can also have a series of displacements. In the previous example of the pacing professor, the individual displacements are 2 m and −4 −4 m, giving a total displacement of −2 m. We define total displacement Δ x Total Δ x Total , as the sum of the individual displacements , and express this mathematically with the equation

where Δ x i Δ x i are the individual displacements. In the earlier example,

The total displacement is 2 − 4 = −2 m along the x -axis. It is also useful to calculate the magnitude of the displacement, or its size. The magnitude of the displacement is always positive. This is the absolute value of the displacement, because displacement is a vector and cannot have a negative value of magnitude. In our example, the magnitude of the total displacement is 2 m, whereas the magnitudes of the individual displacements are 2 m and 4 m.

The magnitude of the total displacement should not be confused with the distance traveled. Distance traveled x Total x Total , is the total length of the path traveled between two positions. In the previous problem, the distance traveled is the sum of the magnitudes of the individual displacements:

Average Velocity

To calculate the other physical quantities in kinematics we must introduce the time variable. The time variable allows us not only to state where the object is (its position) during its motion, but also how fast it is moving. How fast an object is moving is given by the rate at which the position changes with time.

For each position x i x i , we assign a particular time t i t i . If the details of the motion at each instant are not important, the rate is usually expressed as the average velocity v – v – . This vector quantity is simply the total displacement between two points divided by the time taken to travel between them. The time taken to travel between two points is called the elapsed time Δ t Δ t .

If x 1 x 1 and x 2 x 2 are the positions of an object at times t 1 t 1 and t 2 t 2 , respectively, then

It is important to note that the average velocity is a vector and can be negative, depending on positions x 1 x 1 and x 2 x 2 .

Example 3.1

Delivering flyers.

  • What is Jill’s total displacement to the point where she stops to rest?
  • What is the magnitude of the final displacement?
  • What is the average velocity during her entire trip?
  • What is the total distance traveled?
  • Make a graph of position versus time.

A sketch of Jill’s movements is shown in Figure 3.4 .

  • From the above table, the total displacement is ∑ Δ x i = 0.5 − 0.5 + 1.0 − 1.75 km = −0.75 km . ∑ Δ x i = 0.5 − 0.5 + 1.0 − 1.75 km = −0.75 km .
  • The magnitude of the total displacement is | −0.75 | km = 0.75 km | −0.75 | km = 0.75 km .
  • Average velocity = Total displacement Elapsed time = v – = −0.75 km 58 min = −0.013 km/min Average velocity = Total displacement Elapsed time = v – = −0.75 km 58 min = −0.013 km/min
  • The total distance traveled (sum of magnitudes of individual displacements) is x Total = ∑ | Δ x i | = 0.5 + 0.5 + 1.0 + 1.75 km = 3.75 km x Total = ∑ | Δ x i | = 0.5 + 0.5 + 1.0 + 1.75 km = 3.75 km .


Check your understanding 3.1.

A cyclist rides 3 km west and then turns around and rides 2 km east. (a) What is their displacement? (b) What is the distance traveled? (c) What is the magnitude of their displacement?

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Access for free at https://openstax.org/books/university-physics-volume-1/pages/1-introduction
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  • Publisher/website: OpenStax
  • Book title: University Physics Volume 1
  • Publication date: Sep 19, 2016
  • Location: Houston, Texas
  • Book URL: https://openstax.org/books/university-physics-volume-1/pages/1-introduction
  • Section URL: https://openstax.org/books/university-physics-volume-1/pages/3-1-position-displacement-and-average-velocity

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  • Physics Formulas

Displacement Formula

Displacement is calculated as the shortest distance between starting and final point which prefers straight-line path over curved paths.

Suppose a body is moving in two different directions x and y then Resultant Displacement will be

displacement formula

It gives the shortcut paths for the given original paths.

Displacement Formula 2

u = Initial velocity

v = final velocity

a = acceleration

t = time taken.

Solved Examples

Problem 1: The path distance from the garden to a school is 5 m west and then 4 m south. A builder wants to build a short distance path for it. Find the displacement length of the shortest path.

Given: Distance to the west x = 5 m

Distance to the south y = 4 m.

Distance to the south

Displacement is given by

displacement formula 3

s = 6.403 m.

The builder can build a path for displacement length of 6.7 m.

Question 2:A girl walks from the corridor to the gate she moves 3 m to the north opposite to her house then takes a left turn and walks for 5 m, then she takes right turn and moves for 6 m and reaches the gate. What is the displacement, magnitude, and distance covered by her?

Total distance travelled d = 3 m + 5 m + 6 m = 14 m.

A magnitude of the displacement can be obtained by visualizing the walking. The actual path from A to B as 3 m then from B to D as 5 m and finally from D to E as 6 m.

Total distance travelled

So, the magnitude of the resultant displacement is

From figure AC = AB + BC = 3 m + 6 m = 9 m

BD = CE = 5 m

|S| =√92+52 = 10.29 m.

The direction of Resultant displacement is South East.

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How to Calculate Displacement

Last Updated: September 27, 2023 Fact Checked

This article was reviewed by Grace Imson, MA . Grace Imson is a math teacher with over 40 years of teaching experience. Grace is currently a math instructor at the City College of San Francisco and was previously in the Math Department at Saint Louis University. She has taught math at the elementary, middle, high school, and college levels. She has an MA in Education, specializing in Administration and Supervision from Saint Louis University. There are 11 references cited in this article, which can be found at the bottom of the page. This article has been fact-checked, ensuring the accuracy of any cited facts and confirming the authority of its sources. This article has been viewed 464,172 times.

Displacement in physics refers to on object's change in position. When you calculate displacement, you measure how "out of place" on object is based on its initial location and its final location. The formula you use for calculating displacement will depend on variables that are provided to you in a given problem. Follow these steps to calculate displacement.

Calculating Resultant Displacement

Step 1 Use the resultant displacement formula when units of distance are used to specify your initial  and final location.

  • The resultant displacement formula is written as: S = √x²+y² . "S" stands for displacement. X is the first direction that the object is traveling and Y is the second direction that the object is traveling. If your object only travels in one direction, then Y = 0.
  • An object can only travel in two directions maximum, since moving along the north/south or east/west axes will be considered neutral movement.

Step 2 Connect the points based on order of movement and label them from A-Z.

  • Also remember to connect your starting point to your end point using a straight line. This is the displacement we will be calculating.
  • For example, if an object travels east 300 feet and north 400 feet, it will form a right triangle. AB will form the first leg of the triangle and BC will form the second leg. AC will form the hypotenuse of the triangle, and its value will be the amount of the object's displacement. In this example, the two directions are "east" and "north."

Step 3 Input the direction values for x² and y².

  • For example, x = 300 and y = 400. Your formula should look like this: S = √300² + 400².

Step 4 Compute the formula...

  • For example: S = √90000 + 160000. S = √250000. S = 500. You now know that displacement is equal to 500 feet.

When Velocity and Time Values are Specified

Step 1 Use this formula when the problem specifies the object's velocity and time taken.

  • In this case, the formula would be: S = 1/2(u + v)t. U = the object's initial velocity, or how fast it started going in a certain direction. V = the object's final velocity, or how fast it was going at its last location. T = the time the object took to get there. [5] X Research source
  • For example: A car is traveling down the road for 45 seconds (time taken). The car turned west at 20 m/s (initial velocity) and by the end of the street, it was traveling at 23 m/s (final velocity).Calculate the displacement based on these factors.

Step 2 Input the values for velocity and time into their respective variables.

  • Your formula will look like this: S = 1/2(20 + 23)45.

Step 3 Calculate the formula once you have input the values into their correct places.

  • For this formula, it's okay if you accidentally switch initial and final velocity. Since you will be adding these numbers first, it doesn't matter where they are in the parentheses. For other formulas, however, switching initial with final velocity will give you a different displacement value.
  • Your formula will look like this: S = 1/2(43)45. First divide 43 by 2, which will give you 21.5 Then multiply 21.5 by 45, which should give you 967.5 meters. 967.5 is your displacement value, or how far your car is from its original spot.

When Initial Velocity, Acceleration, and Time Values are Specified

Step 1 Employ a modified formula when acceleration is specified along with initial velocity and time.

  • The formula for this problem is as follows: S = ut + 1/2at² . "U" still represents initial velocity; "A" is the object's acceleration, or how fast its velocity begins to change. "T" can either mean the total time taken, or it can be a certain amount of time an object accelerated for. Either way, it will be identified by units of time such as seconds, hours, etc. [8] X Research source
  • Say a car traveling at 25 m/s (initial velocity) begins accelerating at 3 m/s2 (acceleration) for 4 seconds (time). What is the displacement of the car after 4 seconds? [9] X Research source

Step 2 Plug in the values where they belong in the formula.

  • Based on the example data above, your formula should look like this: S = 25(4) + 1/2(3)4². It will help if you add parentheses around your acceleration and time values to help you keep the numbers separate.

Step 3 Calculate displacement by performing the necessary order of operations.

  • Let's revisit the formula: S = 25(4) + 1/2(3)4². First, square 4, which gives you 16. Then multiply 16 by 3, which gives you 48; also multiply 25 by 4, giving you 100. Divide 48 by 2, which is 24. Your equation should now look like this: S = 100 + 24. Once you add the two values together, displacement will equal 124 meters. [12] X Research source

Calculating Angular Displacement

Step 1 Find angular displacement when an object travels in a curved path.

  • Think of a girl sitting on a merry-go-round. As she spins along the outside of the ride, she will travel in a curved path. Angular displacement seeks to measure the shortest distance between the initial location and final location when an object isn't moving in a straight line.
  • The formula for angular displacement is: θ = S/r , where "S" stands for linear displacement, "r" stands for radius, and "θ" represents angular displacement. Linear displacement is how far an object travels along an arc. Radius is the distance an object is from the center of a circle. Angular displacement is the value we are looking for. [13] X Research source

Step 2 Input the linear displacement and radius values into the equation.

  • Here is a sample problem: A girl rides a merry-go-round. Her seat is at a distance of 1 meter from the center (radius). If the girl moves along an arc length of 1.5 meters (linear displacement), what is her angular displacement?
  • Your equation should look like this: θ = 1.5/1.

Step 3 Divide linear displacement by the radius.

  • After dividing 1.5 by 1, you are left with 1.5. The angular displacement of the girl is 1.5 radians.
  • Since angular displacement calculates how much an object has rotated from its original position, it will need to be measured as an angle, not a distance. Radians are units used to measure angles.

Understanding Displacement

Step 1 Know that

  • Distance is what is known as a "scalar quantity." It refers to how much ground an object has covered without considering the direction the object is traveling. [15] X Research source
  • For example, if you walk 2 feet east, 2 feet south, 2 feet west, and then 2 feet north, you will be back in your original position. Though you will have traveled a total distance of 8 feet, you will have been displaced 0 feet because your final location is the same as your original location (your path resembles that of a box). [16] X Research source

Step 2 Understand that displacement is the difference between two locations.

  • Displacement is called a "vector quantity" and refers to an object's change in position with regards to the direction an object is moving.
  • Say you head east for 5 feet. If you go back west 5 feet, you will be traveling in the opposite direction of your original location. Even though you will have walked 10 feet total, you won't have changed your position; your displacement then is 0 feet.

Step 3 Remember the words

  • Picture a football coach pacing back and forth along the sidelines. [18] X Research source As he yells things at his players, he will have moved from left to right multiple times. If you watch him the entire time he is moving from left to right, you are observing the total distance he is traveling. But, say the coach stops to talk to the quarterback on the sidelines. If he is in a spot that is different than before he started pacing, you are looking at the coach's displacement. [19] X Research source

Step 4 Know that displacement is measured using a straight line, not a curved path.

  • A curved path will lead you from your initial location to your final location, but it is not the shortest path. To help you visualize this, imagine you are walking in a straight line and encounter a pillar. You can't walk through this pillar, so you walk around it. Though you will end up in the same position as if you walked through the pillar, you will have needed to take extra steps to get to your destination.
  • Though displacement prefers a straight line, know that you can measure the displacement of an object that is traveling in a curved path. This is called "angular displacement," and it can be calculated by finding the straightest path that leads from initial location to final location.

Step 5 Understand that displacement can be a negative value, unlike distance.

  • For example, say you walked 5 feet east and then 3 feet west. Though technically you are still 2 feet from your original location, your displacement would be -2 because you are moving in the opposite direction. Your distance will always be a positive value because you can't "un-travel" amount of feet, miles, etc.
  • Negative displacement does not mean that the displacement is getting decreased. It simply means that the displacement is taking an opposite direction.

Step 6 Realize that sometimes distance and displacement values can be the same.

  • This only applies to when you travel to one location from your initial location in a straight line. For example, say you live in San Francisco, California and land a new job in Las Vegas, Nevada. You need to move out to Las Vegas in order to be closer to your job. If you take a plane that flies straight from San Francisco to Las Vegas, you will have traveled 417 miles (671 km) and will be displaced 417 miles (671 km).
  • If you take a car from San Francisco to Las Vegas, however, you will be displaced 417 miles (671 km) but will have traveled 563 miles (906 km). [22] X Research source Since driving usually involves changing directions (east on this highway, west on that highway), you will have traveled farther than the shortest distance between the two cities.

Community Q&A

Community Answer

Video . By using this service, some information may be shared with YouTube.

  • For ship displacement, you can carry this process further to figure out how low the ship has to sit in the water. The ship will sit sufficiently low in the water such that the weight of water displaced by the volume of the ship that has pushed down into the water equals the weight of the ship. Thanks Helpful 2 Not Helpful 4

how to solve displacement problems in physics

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  • ↑ https://www.ck12.org/calculus/resultant-applications/lesson/Resultant-of-Two-Displacements-TRIG/
  • ↑ https://www.physicsclassroom.com/class/vectors/Lesson-1/Component-Addition
  • ↑ https://www.calculatorsoup.com/calculators/physics/displacement_v_t.php
  • ↑ http://www.calculatorsoup.com/calculators/physics/displacement_v_t.php
  • ↑ https://www.calculatorsoup.com/calculators/physics/displacement_v_a_t.php
  • ↑ http://www.calculatorsoup.com/calculators/physics/displacement_v_a_t.php
  • ↑ https://pressbooks.bccampus.ca/humanbiomechanics/chapter/6-1-rotation-angle-and-angular-velocity-2/
  • ↑ https://www.omnicalculator.com/physics/angular-displacement
  • ↑ http://www.physicsclassroom.com/class/1DKin/Lesson-1/Distance-and-Displacement
  • ↑ https://www.khanacademy.org/science/physics/one-dimensional-motion/displacement-velocity-time/a/what-is-displacement
  • ↑ http://www.distance-cities.com/distance-las-vegas-nv-to-san-francisco-ca

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4 Motion in Two and Three Dimensions

4.1 displacement and velocity vectors, learning objectives.

By the end of this section, you will be able to:

  • Calculate position vectors in a multidimensional displacement problem.
  • Solve for the displacement in two or three dimensions.
  • Calculate the velocity vector given the position vector as a function of time.
  • Calculate the average velocity in multiple dimensions.

Displacement and velocity in two or three dimensions are straightforward extensions of the one-dimensional definitions. However, now they are vector quantities, so calculations with them have to follow the rules of vector algebra, not scalar algebra.

Displacement Vector

To describe motion in two and three dimensions, we must first establish a coordinate system and a convention for the axes. We generally use the coordinates x , y , and z to locate a particle at point P ( x , y , z ) in three dimensions. If the particle is moving, the variables x , y , and z are functions of time ( t ):

The position vector from the origin of the coordinate system to point P is [latex] \overset{\to }{r}(t). [/latex] In unit vector notation, introduced in Coordinate Systems and Components of a Vector , [latex] \overset{\to }{r}(t) [/latex] is

(Figure) shows the coordinate system and the vector to point P , where a particle could be located at a particular time t . Note the orientation of the x , y , and z axes. This orientation is called a right-handed coordinate system ( Coordinate Systems and Components of a Vector ) and it is used throughout the chapter.

An x y z coordinate system is shown, with positive x out of the page, positive y to the right, and positive z up. A point P, with coordinates x of t, y of t, and z of t is shown. All of P’s coordinates are positive. The vector r of t from the origin to P is also shown as a purple arrow. The coordinates x of t, y of t and z of t are shown as dashed lines. X of t is a segment in the x y plane, parallel to the x axis, y of t is a segment in the x y plane, parallel to the y axis, and z of t is a segment parallel to the z axis.

Figure 4.2 A three-dimensional coordinate system with a particle at position P(x(t), y(t), z(t)).

With our definition of the position of a particle in three-dimensional space, we can formulate the three-dimensional displacement. (Figure) shows a particle at time [latex] {t}_{1} [/latex] located at [latex] {P}_{1} [/latex] with position vector [latex] \overset{\to }{r}({t}_{1}). [/latex] At a later time [latex] {t}_{2}, [/latex] the particle is located at [latex] {P}_{2} [/latex] with position vector [latex] \overset{\to }{r}({t}_{2}) [/latex]. The displacement vector [latex] \text{Δ}\overset{\to }{r} [/latex] is found by subtracting [latex] \overset{\to }{r}({t}_{1}) [/latex] from [latex] \overset{\to }{r}({t}_{2})\text{ }: [/latex]

Vector addition is discussed in Vectors . Note that this is the same operation we did in one dimension, but now the vectors are in three-dimensional space.

An x y z coordinate system is shown, with positive x out of the page, positive y to the right, and positive z up. Two points, P 1 and P 2 are shown. The vector r of t 1 from the origin to P 1 and the vector r of t 2 from the origin to P 2 are shown as purple arrows. The vector delta r is shown as a purple arrow whose tail is at P 1 and head at P 2.

Figure 4.3 The displacement [latex] \text{Δ}\overset{\to }{r}=\overset{\to }{r}({t}_{2})-\overset{\to }{r}({t}_{1}) [/latex] is the vector from [latex] {P}_{1} [/latex] to [latex] {P}_{2} [/latex].

The following examples illustrate the concept of displacement in multiple dimensions.

Polar Orbiting Satellite

A satellite is in a circular polar orbit around Earth at an altitude of 400 km—meaning, it passes directly overhead at the North and South Poles. What is the magnitude and direction of the displacement vector from when it is directly over the North Pole to when it is at [latex] -45\text{°} [/latex] latitude?

We make a picture of the problem to visualize the solution graphically. This will aid in our understanding of the displacement. We then use unit vectors to solve for the displacement.

An x y coordinate system, centered on the earth, is shown. Positive x is to the east and positive y to the north. A blue circle larger than and concentric with the earth is shown. Vector r of t 1 is an orange arrow from the origin to the location where the blue circle crosses the y axis (90 degrees counter clockwise from the positive x axis.) Vector r of t 2 is an orange arrow from the origin to the location on the blue circle at minus 45 degrees. Delta r vector is shown as a purple arrow pointing down and to the right, starting at the head of vector r of t 1 and ending at the head of vector r of t 2.

Figure 4.4 Two position vectors are drawn from the center of Earth, which is the origin of the coordinate system, with the y-axis as north and the x-axis as east. The vector between them is the displacement of the satellite.

In unit vector notation, the position vectors are

Evaluating the sine and cosine, we have

Now we can find [latex] \text{Δ}\overset{\to }{r} [/latex], the displacement of the satellite:

The magnitude of the displacement is [latex] |\text{Δ}\overset{\to }{r}|=\sqrt{{(4787)}^{2}+{(-11,557)}^{2}}=12,509\,\text{km}. [/latex] The angle the displacement makes with the x- axis is [latex] \theta ={\text{tan}}^{-1}(\frac{-11,557}{4787})=-67.5\text{°}. [/latex]


Plotting the displacement gives information and meaning to the unit vector solution to the problem. When plotting the displacement, we need to include its components as well as its magnitude and the angle it makes with a chosen axis—in this case, the x -axis ( (Figure) ).

An x y coordinate system is shown. Positive x is to the east and positive y to the north. Vector delta r sub x points east and has magnitude 4797 kilometers. Vector delta r sub y points south and has magnitude 11,557 kilometers. Vector delta r points to the southeast, starting at the tail of delta r sub x and ending at the head of delta r sub y and has magnitude 12,509 kilometers.

Figure 4.5 Displacement vector with components, angle, and magnitude.

Note that the satellite took a curved path along its circular orbit to get from its initial position to its final position in this example. It also could have traveled 4787 km east, then 11,557 km south to arrive at the same location. Both of these paths are longer than the length of the displacement vector. In fact, the displacement vector gives the shortest path between two points in one, two, or three dimensions.

Many applications in physics can have a series of displacements, as discussed in the previous chapter. The total displacement is the sum of the individual displacements, only this time, we need to be careful, because we are adding vectors. We illustrate this concept with an example of Brownian motion.

Brownian Motion

Brownian motion is a chaotic random motion of particles suspended in a fluid, resulting from collisions with the molecules of the fluid. This motion is three-dimensional. The displacements in numerical order of a particle undergoing Brownian motion could look like the following, in micrometers ( (Figure) ):

What is the total displacement of the particle from the origin?

An x y z coordinate system with distances measured in micrometers and ranging from -10 to +10 micrometers is shown. The displacements delta r sub 1 equals 2 I hat plus j hat plus 2 k hat, delta r sub 2 equals -1 I hat plus 3 k hat, and delta r sub 3 equals -3 I hat plus j hat plus 2 k hat are shown as blue line segments. Vector r 1 hat starts at the origin. Each subsequent displacement starts where the previous one ends. Vector delta r total is shown as a red line starting at the origin and ending at the end of vector delta r 4. Delta r total equals 2 I hat plus 0 y hat plus 9 k hat.

Figure 4.6 Trajectory of a particle undergoing random displacements of Brownian motion. The total displacement is shown in red.

Show Answer

[latex] \begin{array}{cc}\hfill \text{Δ}{\overset{\to }{r}}_{\text{Total}}& =\sum \text{Δ}{\overset{\to }{r}}_{i}=\text{Δ}{\overset{\to }{r}}_{1}+\text{Δ}{\overset{\to }{r}}_{2}+\text{Δ}{\overset{\to }{r}}_{3}+\text{Δ}{\overset{\to }{r}}_{4}\hfill \\ & =(2.0-1.0+4.0-3.0)\hat{i}+(1.0+0-2.0+1.0)\hat{j}+(3.0+3.0+1.0+2.0)\hat{k}\hfill \\ & =2.0\hat{i}+0\hat{j}+9.0\hat{k}\mu \text{m}.\hfill \end{array} [/latex] To complete the solution, we express the displacement as a magnitude and direction,

[latex] |\text{Δ}{\overset{\to }{r}}_{\text{Total}}|=\sqrt{{2.0}^{2}+{0}^{2}+{9.0}^{2}}=9.2\,\mu \text{m,}\quad \theta ={\text{tan}}^{-1}(\frac{9}{2})=77\text{°}, [/latex] with respect to the x-axis in the xz-plane.

From the figure we can see the magnitude of the total displacement is less than the sum of the magnitudes of the individual displacements.

Velocity Vector

In the previous chapter we found the instantaneous velocity by calculating the derivative of the position function with respect to time. We can do the same operation in two and three dimensions, but we use vectors. The instantaneous velocity vector is now

Let’s look at the relative orientation of the position vector and velocity vector graphically. In (Figure) we show the vectors [latex] \overset{\to }{r}(t) [/latex] and [latex] \overset{\to }{r}(t+\text{Δ}t), [/latex] which give the position of a particle moving along a path represented by the gray line. As [latex] \text{Δ}t [/latex] goes to zero, the velocity vector, given by (Figure) , becomes tangent to the path of the particle at time t .

Vectors r of t and r of t plus delta t are shown as red arrows in x y coordinate system. Both vectors start at the origin. Vector delta r points from the head of vector r of t to the head of vector r of t plus delta t.

Figure 4.7 A particle moves along a path given by the gray line. In the limit as [latex] \text{Δ}t [/latex] approaches zero, the velocity vector becomes tangent to the path of the particle.

(Equation) can also be written in terms of the components of [latex] \overset{\to }{v}(t). [/latex] Since

we can write

If only the average velocity is of concern, we have the vector equivalent of the one-dimensional average velocity for two and three dimensions:

Calculating the Velocity Vector

The position function of a particle is [latex] \overset{\to }{r}(t)=2.0{t}^{2}\hat{i}+(2.0+3.0t)\hat{j}+5.0t\hat{k}\text{m}. [/latex] (a) What is the instantaneous velocity and speed at t = 2.0 s? (b) What is the average velocity between 1.0 s and 3.0 s?

Using (Figure) and (Figure) , and taking the derivative of the position function with respect to time, we find

(b) From (Figure) ,

We see the average velocity is the same as the instantaneous velocity at t = 2.0 s, as a result of the velocity function being linear. This need not be the case in general. In fact, most of the time, instantaneous and average velocities are not the same.

Check Your Understanding

The position function of a particle is [latex] \overset{\to }{r}(t)=3.0{t}^{3}\hat{i}+4.0\hat{j}. [/latex] (a) What is the instantaneous velocity at t = 3 s? (b) Is the average velocity between 2 s and 4 s equal to the instantaneous velocity at t = 3 s?

(a) Taking the derivative with respect to time of the position function, we have [latex] \overset{\to }{v}(t)=9.0{t}^{2}\hat{i}\text{and}\overset{\to }{v}\text{(3.0s)}=81.0\hat{i}\text{m/s}. [/latex] (b) Since the velocity function is nonlinear, we suspect the average velocity is not equal to the instantaneous velocity. We check this and find

[latex] {\overset{\to }{v}}_{\text{avg}}=\frac{\overset{\to }{r}({t}_{2})-\overset{\to }{r}({t}_{1})}{{t}_{2}-{t}_{1}}=\frac{\overset{\to }{r}(4.0\,\text{s})-\overset{\to }{r}(2.0\,\text{s})}{4.0\,\text{s}-2.0\,\text{s}}=\frac{(144.0\hat{i}-36.0\hat{i})\,\text{m}}{2.0\,\text{s}}=54.0\hat{i}\text{m/s}, [/latex]

which is different from [latex] \overset{\to }{v}\text{(3.0s)}=81.0\hat{i}\text{m/s}. [/latex]

The Independence of Perpendicular Motions

When we look at the three-dimensional equations for position and velocity written in unit vector notation, (Figure) and (Figure) , we see the components of these equations are separate and unique functions of time that do not depend on one another. Motion along the x direction has no part of its motion along the y and z directions, and similarly for the other two coordinate axes. Thus, the motion of an object in two or three dimensions can be divided into separate, independent motions along the perpendicular axes of the coordinate system in which the motion takes place.

To illustrate this concept with respect to displacement, consider a woman walking from point A to point B in a city with square blocks. The woman taking the path from A to B may walk east for so many blocks and then north (two perpendicular directions) for another set of blocks to arrive at B . How far she walks east is affected only by her motion eastward. Similarly, how far she walks north is affected only by her motion northward.

In the kinematic description of motion, we are able to treat the horizontal and vertical components of motion separately. In many cases, motion in the horizontal direction does not affect motion in the vertical direction, and vice versa.

An example illustrating the independence of vertical and horizontal motions is given by two baseballs. One baseball is dropped from rest. At the same instant, another is thrown horizontally from the same height and it follows a curved path. A stroboscope captures the positions of the balls at fixed time intervals as they fall ( (Figure) ).

Two identical balls are illustrated at 5 locations at equal time intervals. The balls start at the same vertical position. Green arrows represent the horizontal velocities and purple arrows represent the vertical velocities at each position. The ball on the right has an initial horizontal velocity whereas the ball on the left has no horizontal velocity. The horizontal motion is constant horizontal velocity at all times for both balls. The vertical motion is constant vertical acceleration. Each ball’s vertical velocity is increasing in magnitude and pointing down. At each instant in time, both balls have identical vertical positions and vertical velocities.

Figure 4.8 A diagram of the motions of two identical balls: one falls from rest and the other has an initial horizontal velocity. Each subsequent position is an equal time interval. Arrows represent the horizontal and vertical velocities at each position. The ball on the right has an initial horizontal velocity whereas the ball on the left has no horizontal velocity. Despite the difference in horizontal velocities, the vertical velocities and positions are identical for both balls, which shows the vertical and horizontal motions are independent.

It is remarkable that for each flash of the strobe, the vertical positions of the two balls are the same. This similarity implies vertical motion is independent of whether the ball is moving horizontally. (Assuming no air resistance, the vertical motion of a falling object is influenced by gravity only, not by any horizontal forces.) Careful examination of the ball thrown horizontally shows it travels the same horizontal distance between flashes. This is because there are no additional forces on the ball in the horizontal direction after it is thrown. This result means horizontal velocity is constant and is affected neither by vertical motion nor by gravity (which is vertical). Note this case is true for ideal conditions only. In the real world, air resistance affects the speed of the balls in both directions.

The two-dimensional curved path of the horizontally thrown ball is composed of two independent one-dimensional motions (horizontal and vertical). The key to analyzing such motion, called projectile motion , is to resolve it into motions along perpendicular directions. Resolving two-dimensional motion into perpendicular components is possible because the components are independent.

  • The position function [latex] \overset{\to }{r}(t) [/latex] gives the position as a function of time of a particle moving in two or three dimensions. Graphically, it is a vector from the origin of a chosen coordinate system to the point where the particle is located at a specific time.
  • The displacement vector [latex] \text{Δ}\overset{\to }{r} [/latex] gives the shortest distance between any two points on the trajectory of a particle in two or three dimensions.
  • Instantaneous velocity gives the speed and direction of a particle at a specific time on its trajectory in two or three dimensions, and is a vector in two and three dimensions.
  • The velocity vector is tangent to the trajectory of the particle.
  • Displacement [latex] \overset{\to }{r}(t) [/latex] can be written as a vector sum of the one-dimensional displacements [latex] \overset{\to }{x}(t),\overset{\to }{y}(t),\overset{\to }{z}(t) [/latex] along the x , y , and z directions.
  • Velocity [latex] \overset{\to }{v}(t) [/latex] can be written as a vector sum of the one-dimensional velocities [latex] {v}_{x}(t),{v}_{y}(t),{v}_{z}(t) [/latex] along the x , y , and z directions.
  • Motion in any given direction is independent of motion in a perpendicular direction.

Conceptual Questions

What form does the trajectory of a particle have if the distance from any point A to point B is equal to the magnitude of the displacement from A to B ?

straight line

Give an example of a trajectory in two or three dimensions caused by independent perpendicular motions.

If the instantaneous velocity is zero, what can be said about the slope of the position function?

The slope must be zero because the velocity vector is tangent to the graph of the position function.

The coordinates of a particle in a rectangular coordinate system are (1.0, –4.0, 6.0). What is the position vector of the particle?

[latex] \overset{\to }{r}=1.0\hat{i}-4.0\hat{j}+6.0\hat{k} [/latex]

The position of a particle changes from [latex] {\overset{\to }{r}}_{1}=(2.0\text{​}\hat{i}+3.0\hat{j})\text{cm} [/latex] to [latex] {\overset{\to }{r}}_{2}=(-4.0\hat{i}+3.0\hat{j})\,\text{cm}. [/latex] What is the particle’s displacement?

The 18th hole at Pebble Beach Golf Course is a dogleg to the left of length 496.0 m. The fairway off the tee is taken to be the x direction. A golfer hits his tee shot a distance of 300.0 m, corresponding to a displacement [latex] \text{Δ}{\overset{\to }{r}}_{1}=300.0\,\text{m}\hat{i}, [/latex] and hits his second shot 189.0 m with a displacement [latex] \text{Δ}{\overset{\to }{r}}_{2}=172.0\,\text{m}\hat{i}+80.3\,\text{m}\hat{j}. [/latex] What is the final displacement of the golf ball from the tee?

[latex] \text{Δ}{\overset{\to }{r}}_{\text{Total}}=472.0\,\text{m}\hat{i}+80.3\,\text{m}\hat{j} [/latex]

A bird flies straight northeast a distance of 95.0 km for 3.0 h. With the x -axis due east and the y -axis due north, what is the displacement in unit vector notation for the bird? What is the average velocity for the trip?

A cyclist rides 5.0 km due east, then 10.0 km [latex] 20\text{°} [/latex] west of north. From this point she rides 8.0 km due west. What is the final displacement from where the cyclist started?

[latex] \text{Sum of displacements}=-6.4\,\text{km}\hat{i}+9.4\,\text{km}\hat{j} [/latex]

New York Rangers defenseman Daniel Girardi stands at the goal and passes a hockey puck 20 m and [latex] 45\text{°} [/latex] from straight down the ice to left wing Chris Kreider waiting at the blue line. Kreider waits for Girardi to reach the blue line and passes the puck directly across the ice to him 10 m away. What is the final displacement of the puck? See the following figure.

An illustration of the situation described in the problem. The goal and the two ice hockey players are drawn as viewed from above. The goal and Girardi are at the origin of an x y coordinate system. A gray arrow representing 20 meters at 45 degrees from the positive x direction is shown, with Kreider drawn near the tip of the arrow. A blue line, parallel to the x axis, is also drawn at the tip of this arrow. A second gray arrow is shown starting at the Kreider’s location, pointing horizontally to the left, and representing a distance of 10 meters. A dark blue arrow is drawn from the goal at the origin to the tip of the second, 10 meter, gray arrow.

The position of a particle is [latex] \overset{\to }{r}(t)=4.0{t}^{2}\hat{i}-3.0\hat{j}+2.0{t}^{3}\hat{k}\text{m}. [/latex] (a) What is the velocity of the particle at 0 s and at [latex] 1.0 [/latex] s? (b) What is the average velocity between 0 s and [latex] 1.0 [/latex] s?

a. [latex] \overset{\to }{v}(t)=8.0t\hat{i}+6.0{t}^{2}\hat{k},\enspace\overset{\to }{v}(0)=0,\enspace\overset{\to }{v}(1.0)=8.0\hat{i}+6.0\hat{k}\text{m/s} [/latex],

b. [latex] {\overset{\to }{v}}_{\text{avg}}=4.0\text{​}\hat{i}+2.0\hat{k}\,\text{m/s} [/latex]

Clay Matthews, a linebacker for the Green Bay Packers, can reach a speed of 10.0 m/s. At the start of a play, Matthews runs downfield at [latex] 45\text{°} [/latex] with respect to the 50-yard line and covers 8.0 m in 1 s. He then runs straight down the field at [latex] 90\text{°} [/latex] with respect to the 50-yard line for 12 m, with an elapsed time of 1.2 s. (a) What is Matthews’ final displacement from the start of the play? (b) What is his average velocity?

The F-35B Lighting II is a short-takeoff and vertical landing fighter jet. If it does a vertical takeoff to 20.00-m height above the ground and then follows a flight path angled at [latex] 30\text{°} [/latex] with respect to the ground for 20.00 km, what is the final displacement?

[latex] \text{Δ}{\overset{\to }{r}}_{1}=20.00\,\text{m}\hat{j},\text{Δ}{\overset{\to }{r}}_{2}=(2.000\,×\,{10}^{4}\text{m})\,(\text{cos}30\text{°}\hat{i}+\text{sin}\,30\text{°}\hat{j}) [/latex]

[latex] \text{Δ}\overset{\to }{r}=1.700\,×\,{10}^{4}\text{m}\hat{i}+1.002\,×\,{10}^{4}\text{m}\hat{j} [/latex]

  • OpenStax University Physics. Authored by : OpenStax CNX. Located at : https://cnx.org/contents/[email protected]:Gofkr9Oy@15 . License : CC BY: Attribution . License Terms : Download for free at http://cnx.org/contents/[email protected]

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