Distance and Displacement in a Plane Revision Notes for HSC SSCE Physics

Distance and Displacement in a Plane

Understanding displacement in two dimensions

In kinematics, it is essential to distinguish between two fundamental quantities: distance and displacement. While these terms might seem similar, they represent different physical concepts, especially when analysing motion in a plane (two dimensions).

Distance is a scalar quantity that measures the total length of the path travelled by an object. It only has magnitude (size) and is always positive.

Displacement is a vector quantity that describes the straight-line distance from an object's starting position to its final position, including the direction of that straight line. Because displacement is a vector, it has both magnitude and direction.

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Key Difference

The crucial distinction is that distance depends on the actual path taken, whereas displacement only depends on the starting and ending positions. The route travelled between these points is irrelevant to the displacement.

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Real-World Application: Multiple Routes, Same Displacement

Consider the example shown above. Cars travelling from Town A to Town B could take different routes along the winding road. One car might travel 170 km, another 230 km (taking a different route), and a third might drive 400 km by going to Town B and back to Town A.

However, all cars at Town B have the same displacement from Town A: approximately 150 km in a north-easterly direction. The car that returned to Town A has zero displacement, even though it travelled 400 km, because it ended at its starting position.

Mathematical representation of displacement

Displacement is mathematically defined as the difference between the final position vector and the initial position vector:

$\vec{s} = \vec{d_f} - \vec{d_i}$

where:

$\vec{s}$ is the displacement vector
$\vec{d_f}$ is the final position vector
$\vec{d_i}$ is the initial position vector

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Position vs. Displacement

It's important to note that a displacement can begin anywhere in space. However, a position vector is always measured relative to the origin of the coordinate system. This is a crucial distinction in problem-solving.

Displacement can also vary with time. The kinematic equation $s = ut + \frac{1}{2}at^2$ allows us to calculate the magnitude of displacement as a function of time (where $u$ is initial velocity, $a$ is acceleration, and $t$ is time). When working in two dimensions, we must account for both the vector nature of displacement and its time dependence.

Adding displacements using the graphical method

When an object undergoes multiple displacements in sequence, we need to find the total or resultant displacement. Since displacements are vectors, we cannot simply add their magnitudes; we must use vector addition techniques.

Head-to-tail method

The head-to-tail method is a graphical technique for adding vectors. To add two displacement vectors $\vec{A}$ and $\vec{B}$ :

Draw the first vector $\vec{A}$ to scale with the correct direction
Place the tail of the second vector $\vec{B}$ at the head (arrow tip) of vector $\vec{A}$
Draw the resultant vector $\vec{R}$ from the tail of $\vec{A}$ to the head of $\vec{B}$

The resultant vector $\vec{R}$ represents the total displacement. This method works for any number of vectors - simply continue placing each subsequent vector head-to-tail.

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Essential for Accurate Results

When using this graphical method, it is essential that all vectors are drawn to the same scale and with accurate angles. A ruler and protractor can be used to measure the magnitude and direction of the resultant.

Parallelogram rule

An alternative graphical method is the parallelogram rule. This method is particularly useful when you want to see both vectors starting from the same point:

Draw both vectors $\vec{A}$ and $\vec{B}$ with their tails at the same point
Complete a parallelogram using these vectors as two adjacent sides
The resultant vector $\vec{R}$ is the diagonal of the parallelogram, running from the common tail to the opposite corner

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Commutative Property of Vector Addition

The parallelogram rule demonstrates that vector addition is commutative: $\vec{A} + \vec{B} = \vec{B} + \vec{A}$ . The order in which you add vectors doesn't affect the final result.

Subtracting vectors

Vector subtraction is performed by adding the negative of a vector. To subtract vector $\vec{B}$ from vector $\vec{A}$ :

$\vec{A} - \vec{B} = \vec{A} + (-\vec{B})$

The negative of a vector, $-\vec{B}$ , is obtained by reversing its direction (flipping the arrow to point the opposite way) while keeping the same magnitude.

Once the second vector is reversed, the head-to-tail method can be applied as usual. The parallelogram rule also works for subtraction, provided you reverse the direction of the second vector before constructing the parallelogram.

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Vector Subtraction is NOT Commutative

Note that vector subtraction is not commutative:

$\vec{A} - \vec{B} = \vec{A} + (-\vec{B}) = (-\vec{B}) + \vec{A}$

but

$\vec{A} - \vec{B} \neq -\vec{A} + \vec{B}$

This is a common mistake to avoid when working with vectors.

Vector subtraction is particularly important when calculating velocity, which is the change in displacement divided by the change in time. Since displacement is a vector, the change in displacement is found by subtracting the initial displacement from the final displacement.

Resolving and adding vectors using components

While graphical methods are useful for visualisation, a mathematical approach using vector components provides more precision and is essential for calculations.

The component method

To add vectors mathematically:

Choose a coordinate system (typically with perpendicular x and y axes, or compass directions like East-West and North-South)
Resolve each vector into its components along these axes
Add all the components in each direction separately
Combine the resulting components to find the magnitude and direction of the resultant vector

For two displacement vectors $\vec{s_1}$ and $\vec{s_2}$ , if the resultant is $\vec{s}$ :

$\vec{s} = \vec{s_1} + \vec{s_2}$

Each vector can be expressed in terms of its components:

$\vec{s} = \vec{s_x} + \vec{s_y}$ $\vec{s_1} = \vec{s_{1,x}} + \vec{s_{1,y}}$ $\vec{s_2} = \vec{s_{2,x}} + \vec{s_{2,y}}$

The components can be added separately:

$\vec{s_x} = \vec{s_{1,x}} + \vec{s_{2,x}}$ $\vec{s_y} = \vec{s_{1,y}} + \vec{s_{2,y}}$

Therefore, the resultant vector can be written as:

$\vec{s} = (\vec{s_{1,x}} + \vec{s_{2,x}}) + (\vec{s_{1,y}} + \vec{s_{2,y}})$

This method works equally well for adding more than two vectors.

Finding magnitude and direction

Once you have the components of the resultant vector, you can find its magnitude using Pythagoras' theorem:

$s = \sqrt{(s_{1,x} + s_{2,x})^2 + (s_{1,y} + s_{2,y})^2}$

The direction can be found using trigonometry. The vectors $\vec{s}$ , $\vec{s_x}$ , and $\vec{s_y}$ form a right-angled triangle, so:

$\tan\theta = \frac{s_y}{s_x}$

where $\theta$ is the angle of the resultant vector relative to the x-axis.

Worked example: Orienteering displacement

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Worked Example: Calculating Resultant Displacement

Let's examine a practical problem that demonstrates the component method for adding displacements.

Problem: Orienteers run two legs of a course. First, they run 600 m S30°E from the origin. Then they run 400 m N60°E. What is their final displacement relative to where they started? (Assume three significant figures for the distances.)

Solution:

First, we draw a diagram showing the two displacement vectors and define our coordinate system. We'll use compass directions, with East as the positive horizontal direction and North as the positive vertical direction.

Let $\vec{s_1}$ be the first displacement (600 m S30°E) and $\vec{s_2}$ be the second displacement (400 m N60°E). The resultant displacement is $\vec{s} = \vec{s_1} + \vec{s_2}$ .

Given data:

$s_1 = 600$ m; $\theta_1 = 30°$
$s_2 = 400$ m; $\theta_2 = 60°$

Step 1: Resolve each displacement into East-West and North-South components

East-West components: $s_{1,\text{East}} = s_1 \sin\theta_1$ $s_{2,\text{East}} = s_2 \sin\theta_2$

North-South components: $s_{1,\text{North}} = -s_1 \cos\theta_1$ (negative because it's southward) $s_{2,\text{North}} = s_2 \cos\theta_2$

Step 2: Substitute the values

Calculation	Explanation
$s_{1,\text{East}} = (600 \text{ m})\sin 30° = 300$ m E	Eastward component of first displacement
$s_{2,\text{East}} = (400 \text{ m})\sin 60° = 346$ m E	Eastward component of second displacement
$s_{1,\text{North}} = -(600 \text{ m})\cos 30° = -520$ m N	Southward component (negative)
$s_{2,\text{North}} = (400 \text{ m})\cos 60° = 200$ m N	Northward component

Step 3: Add the components in each direction

$s_{\text{East}} = s_{1,\text{East}} + s_{2,\text{East}} = 300 \text{ m} + 346 \text{ m} = 646 \text{ m E}$

$s_{\text{North}} = s_{1,\text{North}} + s_{2,\text{North}} = -520 \text{ m} + 200 \text{ m} = -320 \text{ m N}$

The negative value for the North component indicates the net displacement is southward.

Step 4: Find the magnitude of the resultant displacement

Using Pythagoras' theorem:

$s = \sqrt{s_{\text{East}}^2 + s_{\text{North}}^2}$

$s = \sqrt{(646)^2 + (-320)^2} = 721 \text{ m}$

Step 5: Calculate the direction

$\tan\theta = \frac{s_{\text{North}}}{s_{\text{East}}} = \frac{-320 \text{ m}}{646 \text{ m}} = -0.495$

$\theta = -26°$

The negative angle indicates the direction is south of east.

Final Answer:

$\vec{s} = 721 \text{ m E26°S}$

or equivalently:

$\vec{s} = 721 \text{ m S64°E}$

(since $90° - 26° = 64°$ )

This result tells us that the orienteers' final position is 721 metres from their starting point, in a direction 26° south of due east (or 64° east of due south). The actual distance they ran was 1000 m (600 m + 400 m), but their displacement is only 721 m.

Key considerations when working with vectors

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Important Guidelines for Vector Problems

When solving problems involving displacement in a plane:

Always define your coordinate system clearly - whether you're using x-y coordinates or compass directions (N, S, E, W), make sure you specify which direction is positive
Be careful with signs - in compass notation, north and south are opposites, as are east and west. Use positive and negative signs consistently
Distinguish between position and displacement - a position is measured from the origin, while a displacement can start anywhere
Remember that displacement depends on when you measure it - an object's displacement changes throughout its journey
Use appropriate units - always include units (metres, kilometres, etc.) in your calculations and final answer
Express direction clearly - use compass bearings or angles measured from a reference direction

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Key Points to Remember:

Displacement is a vector with both magnitude and direction, representing the straight-line distance from start to finish. Distance is a scalar measuring the total path length travelled.
The path taken doesn't matter for displacement - only the starting and ending positions determine the displacement vector, regardless of the route travelled between them.
Vectors can be added graphically or mathematically - use the head-to-tail method or parallelogram rule for graphical addition, or resolve into components for precise calculations: $\vec{s} = (\vec{s_{1,x}} + \vec{s_{2,x}}) + (\vec{s_{1,y}} + \vec{s_{2,y}})$
Vector subtraction means adding the negative - to subtract $\vec{B}$ from $\vec{A}$ , reverse the direction of $\vec{B}$ and add: $\vec{A} - \vec{B} = \vec{A} + (-\vec{B})$
Use Pythagoras and trigonometry to find resultants - after adding components, find magnitude with $s = \sqrt{s_x^2 + s_y^2}$ and direction with $\tan\theta = \frac{s_y}{s_x}$

Distance and Displacement in a Plane (HSC SSCE Physics): Revision Notes