Position, Distance and Displacement on the Coordinate Plane for SS 1

Position, Distance and Displacement

Performance Objectives

By the end of the lesson, students should be able to:

1. explain the concept of position using the x-y plane and locate the position of an object on a graph using the Cartesian coordinate system;
2. explain the concepts of distance and displacement;
3. differentiate among position, distance and displacement with examples;
4. Measure and calculate the distance and displacement of a body.

Introduction to Motion

In physics, understanding how objects move requires a clear way to describe their location and how that location changes. Concepts like position, distance, and displacement are fundamental tools used to describe motion accurately. These concepts help to quantify movement and provide a basis for more advanced studies in kinematics.

Understanding Position

Position refers to the location of an object relative to a fixed reference point. To describe position accurately, a coordinate system is used. The most common system for describing positions in two dimensions is the Cartesian coordinate system, also known as the x-y coordinate plane.

The Cartesian Coordinate System

The Cartesian coordinate system is a two-dimensional plane formed by two perpendicular number lines: the horizontal x-axis and the vertical y-axis. These axes intersect at a point called the origin (0,0).

Every point on this plane can be uniquely identified by an ordered pair of numbers (x, y), where ‘x’ represents the horizontal distance from the origin along the x-axis, and ‘y’ represents the vertical distance from the origin along the y-axis.

The x-axis values are positive to the right of the origin and negative to the left. The y-axis values are positive above the origin and negative below. This divides the plane into four quadrants.

Locating a Body’s Position

To locate the position of an object on the x-y plane, its coordinates (x, y) are identified. For example, if a car is at position (5, 3), it means it is 5 units along the positive x-axis and 3 units along the positive y-axis from the origin.

Consider an object starting at the origin (0,0). If it moves to a point A, its position is described by the coordinates of A. If it then moves to point B, its new position is described by the coordinates of B.

Understanding Distance

Distance is the total length of the path covered by an object during its motion.

Distance is a scalar quantity, meaning it only has magnitude (a numerical value) and no direction. For example, saying “10 metres” describes a distance, but it does not specify the direction of movement.

The unit of distance in the International System of Units (SI) is the metre (m).

When an object moves from one point to another, the distance covered depends on the actual path taken. If a person walks from their house to a market using a winding road, the distance covered is the length of that entire road.

Understanding Displacement

Displacement is the straight-line distance between an object’s initial position and its final position, along with the direction of that straight line.

Displacement is a vector quantity, meaning it has both magnitude and direction. For example, saying “10 metres East” describes a displacement, as it includes both the magnitude (10 metres) and the direction (East).

The unit of displacement in the International System of Units (SI) is also the metre (m).

Displacement only considers the starting and ending points, not the path taken. If a person walks from their house to a market, their displacement is the length and direction of the straight line connecting the house to the market, regardless of the actual path they followed.

Distinguishing Position, Distance, and Displacement

These three concepts are related but distinct:

1. Position: This describes where an object is at a specific moment in time, relative to a reference point. It is given by coordinates (x, y) in a coordinate system. Position is a vector quantity because it has both magnitude (distance from origin) and direction.
2. Distance: This is the total length of the path travelled by an object. It is a scalar quantity, meaning it only has magnitude. Distance is always positive and increases as an object moves.
3. Displacement: This is the change in an object’s position, represented by a straight line from the starting point to the ending point, including its direction. It is a vector quantity. Displacement can be positive, negative, or zero. If an object returns to its starting point, its displacement is zero, even if it has covered a significant distance.

Consider a person walking around a rectangular field. If they start at one corner, walk along the perimeter, and return to the same corner:

1. Their initial and final position are the same.
2. The distance covered is the total length of the perimeter of the field.
3. Their displacement is zero because their final position is the same as their initial position.

Measuring and Calculating Distance

Measuring Distance on a Graph

To measure distance on a graph, a scale must be established (e.g., 1 cm = 1 metre). The procedure involves:

1. Plotting the path: Draw the path taken by the object on graph paper, marking the starting and ending points, and any intermediate points where the direction of motion changes.
2. Measuring segments: Use a meter rule (or a ruler) to measure the length of each straight-line segment of the path.
3. Summing the lengths: Add the lengths of all segments to find the total distance covered.

For example, if an object moves from point A to B, then B to C, the distance is the sum of the length of line segment AB and the length of line segment BC.

Calculating Distance from Coordinates

When an object moves between two points on a coordinate plane, its distance can be calculated.

1. Movement along an axis: If an object moves horizontally from (x₁, y₁) to (x₂, y₁), the distance is |x₂ – x₁|. If it moves vertically from (x₁, y₁) to (x₁, y₂), the distance is |y₂ – y₁|.
2. Movement along a diagonal path (straight line between two points): The distance ‘d’ between two points (x₁, y₁) and (x₂, y₂) can be calculated using the distance formula, which is derived from the Pythagorean theorem:

$$d = \sqrt{(x_2 – x_1)^2 + (y_2 – y_1)^2}$$

Measuring and Calculating Displacement

Measuring Displacement on a Graph

To measure displacement on a graph, the procedure involves:

1. Plotting initial and final positions: Mark the initial position (start) and the final position (end) of the object on graph paper.
2. Drawing the displacement vector: Draw a straight arrow from the initial position to the final position. This arrow represents the displacement vector.
3. Measuring magnitude: Use a meter rule (or a ruler) to measure the length of this arrow. Convert this length to actual units using the graph’s scale. This is the magnitude of the displacement.
4. Measuring direction: Use a protractor to measure the angle the displacement arrow makes with a reference direction, usually the positive x-axis (East) or North. For example, an angle measured counter-clockwise from the positive x-axis.

Calculating Displacement from Coordinates

To calculate displacement, both its magnitude and direction must be determined.

1. Magnitude: The magnitude of displacement is the straight-line distance between the initial position (x₁, y₁) and the final position (x₂, y₂). This is calculated using the same distance formula as above:

$$|\vec{d}| = \sqrt{(x_2 – x_1)^2 + (y_2 – y_1)^2}$$

Here, $|\vec{d}|$ represents the magnitude of the displacement vector.

2. Direction: The direction of displacement is typically given as an angle (θ) relative to the positive x-axis. This angle can be found using trigonometry:

$$\tan \theta = \frac{\text{change in y}}{\text{change in x}} = \frac{y_2 – y_1}{x_2 – x_1}$$

The angle θ is then found by taking the inverse tangent (arctan) of this ratio. The quadrant of the final position relative to the initial position helps determine the correct angle.

Worked Examples

Example 1: Movement along a straight path

A car starts at point A (2, 1) and drives straight to point B (6, 4) on a coordinate plane where each unit represents 1 kilometre.

1. Calculate the distance covered by the car.
2. Determine the displacement of the car.

Solution:

Initial position (x₁, y₁) = (2, 1)

Final position (x₂, y₂) = (6, 4)

1. Distance covered: Since the car drives straight from A to B, the distance covered is the straight-line distance between A and B.

$$d = \sqrt{(x_2 – x_1)^2 + (y_2 – y_1)^2}$$

$$d = \sqrt{(6 – 2)^2 + (4 – 1)^2}$$

$$d = \sqrt{(4)^2 + (3)^2}$$

$$d = \sqrt{16 + 9}$$

$$d = \sqrt{25}$$

$$d = 5 \text{ km}$$

The distance covered by the car is 5 km.

2. Displacement of the car:

Magnitude: The magnitude of displacement is the same as the straight-line distance calculated above, which is 5 km.

Direction:

$$\tan \theta = \frac{y_2 – y_1}{x_2 – x_1} = \frac{4 – 1}{6 – 2} = \frac{3}{4}$$

$$\theta = \arctan\left(\frac{3}{4}\right)$$

$$\theta \approx 36.87^\circ$$

The displacement of the car is 5 km at an angle of approximately 36.87° from the positive x-axis (or North of East).

Example 2: Movement along a path with turns

A student walks from their home at the origin (0, 0) to a shop at (4, 0), then turns and walks to a friend’s house at (4, 3). Each unit represents 100 metres.

1. Calculate the total distance covered by the student.
2. Determine the student’s displacement from home to the friend’s house.

Solution:

Starting point (Home) = P₀ (0, 0)

Intermediate point (Shop) = P₁ (4, 0)

Final point (Friend’s house) = P₂ (4, 3)

1. Total distance covered:

Distance from Home to Shop (P₀ to P₁):

$$d_1 = |x_1 – x_0| = |4 – 0| = 4 \text{ units}$$

Distance from Shop to Friend’s house (P₁ to P₂):

$$d_2 = |y_2 – y_1| = |3 – 0| = 3 \text{ units}$$

Total distance = d₁ + d₂ = 4 + 3 = 7 units.

Since 1 unit = 100 metres, total distance = 7 × 100 m = 700 m.

The total distance covered by the student is 700 m.

2. Student’s displacement:

Displacement is from the initial position (0, 0) to the final position (4, 3).

Magnitude:

$$|\vec{d}| = \sqrt{(x_2 – x_0)^2 + (y_2 – y_0)^2}$$

$$|\vec{d}| = \sqrt{(4 – 0)^2 + (3 – 0)^2}$$

$$|\vec{d}| = \sqrt{(4)^2 + (3)^2}$$

$$|\vec{d}| = \sqrt{16 + 9}$$

$$|\vec{d}| = \sqrt{25}$$

$$|\vec{d}| = 5 \text{ units}$$

In metres, magnitude = 5 × 100 m = 500 m.

Direction:

$$\tan \theta = \frac{y_2 – y_0}{x_2 – x_0} = \frac{3 – 0}{4 – 0} = \frac{3}{4}$$

$$\theta = \arctan\left(\frac{3}{4}\right)$$

$$\theta \approx 36.87^\circ$$

The student’s displacement is 500 m at an angle of approximately 36.87° from the positive x-axis.