Position, Distance and Displacement

Position in Physics

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Position is the location of an object relative to a chosen reference point called the origin. Position only makes sense with a reference; for instance, saying "the school is 2 km away" is incomplete until you say from where and in what direction.

Position is a vector quantity that describes the exact location of an object in space relative to a reference point called the origin.

 

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Position in One Dimension (a straight line)

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The position is a single number "x", called the (x-coordinate or abscissa) is positive on one side of the origin and negative on the other. Below are the key properties of x:

• Origin: The reference point where x = 0.

• Positive side (x > 0): This indicates the position is to the right (or in the positive direction) of the origin.

• Negative side (x < 0): Indicates the position is to the left (or in the negative direction) of the origin.

• Absolute value (|x|): It represents the exact distance from the origin regardless of direction.

Please read the introduction to physics (definition, branches, importance and scholars of physics) here.

 

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Position in Two Dimensions (the Cartesian x-y plane)

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The plane has a horizontal "x-axis" and a vertical "y-axis" crossing at the "origin O(0, 0)". Any point is written as an ordered pair (x, y). This is shown via the image below:

 

The table below shows the Cartesian x - y plane

Quadrant	x	y
1st	+	+
2nd	-	+
3rd	-	-
4th	+	-

 

• Quadrant I: (+ , +) — Top-right section

• Quadrant II: (- , +) — Top-left section

• Quadrant III: (- , -) — Bottom-left section

• Quadrant IV: (+ , -) — Bottom-right section

 

Euclidean Distance Formula: To find the distance d between two points (x₁, y₁) and (x₂, y₂), use the formula below:

√(x₂ - x₁)² + (y₂ - y₁)²

You can read on fundamental, derived and supplementary quantities and unit in physics (and dimensions) here.

 

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Position Vector

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A position vector is a straight-line vector that starts at a fixed reference point (usually the origin) and points directly to the location of a specific point in space. It has both magnitude and direction.

The position of P(x, y) can be written as a vector from the origin to P:

The equation r = xi + yj represents a position vector in a two-dimensional Cartesian coordinate system. It points from the origin (0,0) to a specific point (x,y) in space.

• r = xi + yj

r is the position vector itself.

x and y are the scalar components (or coordinates) along the horizontal (x) and vertical (y) axes.

i is a unit vector pointing in the positive x-direction (horizontal axis) with a length of 1

j is a unit vector pointing in the positive y-direction (vertical axis) with a length of 1. 

xi and yj: The vector components (or resolved parts) of r in the direction of each axis.

Note: Together, I and j act as direction guides, multiplying the scalar components x and y to show how far along each axis a point or object sits from the origin (0, 0)

Below are some terms to understand with regards to position.

Magnitude (that is, length or how far): This is the straight-line distance from the origin to the point. It is denoted by |r| or simply r, and calculated using the Pythagorean theorem:

• Magnitude: r = √(x² + y²)

 

Direction (that is, which way) of the vector represents the angle of orientation of the vector relative to a fixed reference line, typically the positive x-axis. It is denoted by θ, and calculated using trigonometry (the tangent function):

• Direction: θ = tan⁻¹(y/x), measured from the +x axis (usually in the counterclockwise from the positive x-axis). It tells us exactly which way the vector is pointing (adjust for the quadrant).

 

Worked Example:

• P(3, 4). Find the magnitude and direction.

r = √(9 + 16)

r = 5 m

θ = tan⁻¹(4/3)

θ = 53.1°

Please read on the measurement of basic quantities and importance of measurement in physics here.

 

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Distance

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Distance is the total length of the path an object travels, regardless of direction. It has the following features:

• It is a scalar. This means that it has magnitude only (size) but no direction.

• The SI unit is metre (m). Others include km, cm.

• If an object moves, the distance increases or stays at zero if it does not move. Therefore, the total distance travelled is always positive and never decreases as motion continues. 

• It depends on the path taken.

• It is represented by the variable (d) or (s)

• A car's odometer measures distance.

 

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Displacement

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Displacement is the shortest straight-line distance from the initial position to the final position, in a specified direction. It is the change in position in a given direction. It has the following features:

• It is a vector (that is, has magnitude and direction).

• Its SI unit is metre (m).

• It depends only on the start and end points, not on the path.

• It can be positive, negative or zero. Positive or negative displacement indicates the direction along a coordinate axis (like right/left for x axis or up/down for y axis). It is zero if an object finishes its journey at the exact same point it started.

• The symbol for displacement is Δx or Δs or s or Δr.

 

• In one dimension: 

Δx = x₂ − x₁ (final − initial)

 

• In two dimensions: for a move from A(x₁, y₁) to B(x₂, y₂):

Δr = (x₂ − x₁)i + (y₂ − y₁)j

Magnitude: |Δr| = √[(x₂ − x₁)² + (y₂ − y₁)²]

Direction: θ = tan⁻¹(Δy/Δx)

 

• Displacement Using Pythagoras

You can use the Pythagorean Theorem (a² + b² = c²) to find the straight-line distance between two points by treating that distance as the hypotenuse of a right-angled triangle.

When an object moves along two perpendicular directions (for example east then north), the displacement is the hypotenuse of the right-angled triangle formed.

 

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Distance vs Displacement

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The table below shows the differences between distance and displacement.

Feature	Distance	Displacement
Definition	Total path length	Straight-line change in position
Type / Quantity	Scalar	Vector
Direction	Not needed	Needed
Sign	Always positive	Positive, Negative or Zero
Depends on path?	Yes	No
Over a round trip	Not zero	Zero

 

Key facts

• Distance ≥ |displacement| always.

• They are equal in magnitude only when motion is in a straight line with no change of direction.

• Displacement can be zero while distance is large (a complete lap or round trip).

Displacement is zero: This means that if you finish a complete lap or a round trip, your final position is the same as your starting position, which means your net change in position (displacement) is 0.

Distance is large: Since distance is the total length of the actual path you traveled; and you eventually ended up where you started, you still accumulated a total path length (for example, running a 400-meter track.

Please read on the concept of velocity and acceleration here.

 

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Worked Examples

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Example 1. A boy walks 4 m east, then 3 m north. Find the distance, displacement and direction 

Solution

• Distance = 4 + 3 = 7 m

• Displacement: The image below illustrates the displacement.

(Using the Pythagorean theorem (c² = a² + b²): = (c = √a² + b²)

c = √(4² + 3²)

c= √25  = 5 m

• Direction: = tan θ

tan θ = Opposite/Adjacent = North/East

tan θ = 3/4 = 0.75

θ = 36.9° north of east

 

Example 2: A particle moves from x = +5 m to x = −3 m. Find Δx and the distance covered.

Solution 

• Δx (Displacement) = Final position - Initial position 

Δx = x_j - x_i

Δx = −3 − (+5)

Δx = −8 m (The negative sign indicates that the displacement is in the negative x-direction).

• Distance = |Δx|

• |Δx| = | - 8 m| = 8m

 

Example 3. A particle starts at the origin, moves to +10 m, then returns to +4 m. Find the distance covered and its displacement.

Solution

• The distance include the forward path and return path

Forward path: Moving from the origin (0 m) to (+10) covers a distance of(10 m}

Return path: Moving from (+10 m) back to (+4 m}) covers a distance of |4 - 10| =  6 m (Distance is always positive)

Distance = 10 m + 6 m = 16 m

• Displacement = Final position - Initial position (x_j - x_i)

Initial Position (x_i) = 0 m

Final Position (x_j) = 4 m

Displacement= 4 m − 0 m = +4 m

 

Example 4. A circular track has radius 7 m (π = 22/7). A runner starts at A. Find the full lap distance, displacement and half lap distance, displacement.

Solution

• The full distance covered in one full lap is equal to the circumference (C). C =  2πr.

Full lap: distance = Circumference of the circle (circular track) = 2πr

Full lap distance = 2 × 22/7 × 7 = 44 m

• Displacement = The shortest straight-line distance between the starting position and the final position

Displacement = 0 (The runner finishes exactly where they started (since he/she ran full lap), making the change in position zero.

• Half Lap = 2r = 2 × 7 =14 m (the diameter). Below is the reason for the answer.

After a half lap, the runner ends up at the exact opposite side of the circular track. The shortest path between these two points is a straight line through the center, which is the displacement (or diameter - d) of the circle.

Displacement = 2r (Diameter of the semi circle is the displacement)

2r = 2 × 7 =14 m (the diameter)

 

Example 5: Find the displacement from A(2, 3) to B(8, 11).

Solution 

• Displacement = (x₂ - x₁)i + (y₂ - y₁)j

Initial Point A: (x₁, y₁) = (2, 3)

Final Point B: (x₂, y₂) = (8, 11)

Substitute the coordinates into the formula = (8 -2)i + (11 - 3)j

Displacement= 6i​​​​​​I + 8j = (Δx = 6, Δy = 8)

• Displacement = magnitude and direction

Magnitude: r = √(x² + y²)

Δx = 6, Δy = 8

|Δr| = √(36 + 64)

|Δr| = √100 = 10 units

Direction = tan θ = Opposite/Adjacent = Δy/Δx (OR θ = tan⁻¹(y/x))

tan θ = 8/6 = 1.33

θ = 53.1° above the +x axis

 

Example 6. A ship sails 6 km north, 8 km east, then 6 km south. Find the distance and displacement.

• Distance: It only cares about the total ground covered, regardless of direction.

Distance= 6 + 8 + 6 = 20 km

• Displacement = Net eastward movement, since the two opposite movements cancel each other out completely on the North-South axis.

Displacement = 8 + 6 − 6 = 8, so displacement = 8 km East

You can attempt Len Academy questions and answers in physics here.

 

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Common Mistakes on Distance and Displacement 

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• Treating distance and displacement as the same thing.

• Giving displacement without a direction.

• Forgetting that displacement = final − initial, not the reverse.

• Adding perpendicular displacements directly instead of using Pythagoras.

Please read the next topic (Concept of direction, scalars and vectors) here.

 

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Practice Questions

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Section A: Objective

1. Which of these is a vector quantity? A. Distance B. Speed C. Displacement D. Time

2. A student walks 30 m north then 40 m east. His displacement is _____ A. 10 m B. 50 m C. 70 m D. 1200 m

3. A runner completes one full lap of a 400 m track. His distance and displacement are  _____. A. 400 m, 400 m B. 0, 0 C. 400 m, 0 D. 0, 400 m

4. The magnitude of the position vector of point P(−3, 4) is _____. A. 1 m B. 5 m C. 7 m D. 12 m

5. A body moves from x = 7 m to x = 2 m. Its displacement is _____. A. +9 m B. +5 m C. −5 m D. −9 m

6. Which quantity can be zero even when the object has travelled a long way? A. Distance B. Speed C. Displacement D. Time

7. Point (−2, −5) lies in the _____. A. first quadrant B. second quadrant C. third quadrant D. fourth quadrant

8. A car goes 3 km east then 3 km west. Its distance and displacement are _____. A. 6 km, 0 B. 0, 6 km C. 6 km, 6 km D. 3 km, 3 km

9. Which statement is TRUE? A. Distance can be negative B. Displacement is always equal to distance C. Distance is never less than the magnitude of displacement D. Displacement is a scalar

10. The SI unit of displacement is _____. A. kilometre B. metre C. second D. metre per second

 

Section B: Short Answer

1. Define position and explain why a reference point is necessary.

2. State four differences between distance and displacement.

3. Can distance ever be negative? Can displacement? Explain with examples.

4. Give an example of motion in which the magnitude of displacement equals the distance covered.

5. State the signs of x and y in each of the four quadrants of the Cartesian plane.

6. Explain why a person who returns to the starting point has zero displacement but non-zero distance.

 

Section C: Calculations

1. An object moves from A(1, 2) to B(7, 10). Find (a) the displacement vector, (b) its magnitude, (c) its direction.

2. A boy walks 12 m west then 5 m south. Find (a) the distance covered, (b) his displacement, (c) its direction.

3. A runner moves on a circular track of radius 35 m (π = 22/7). Find the distance and displacement for (a) one full lap, (b) half a lap, (c) a quarter lap.

4. An object moves along O(0, 0) → A(0, 6) → B(8, 6) → C(8, 0). Find (a) the total distance, (b) the displacement from O to B, (c) the displacement from O to C.

5. An aircraft flies 300 km east, 400 km north, then 100 km west. Find the total distance and the magnitude of the final displacement.

6. A particle on a number line starts at x = −4 m, moves to +6 m, then to +1 m. Find (a) the distance travelled, (b) the displacement.

 

Section D: Challenge Questions

1. A rectangular field measures 120 m by 50 m. A man starts at corner A and walks along the edges to the opposite corner C. Find the shortest distance he can walk and his displacement. What is the displacement after he completes one full round of the field?

2. The position vectors of a particle at two times are r₁ = (3i + 4j) m and r₂ = (−2i + 16j) m. Find the displacement vector, its magnitude and its direction.

3. A student claims, "Displacement can sometimes be greater than distance." Is this correct? Justify your answer.