# How to Find Displacement

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Physics can be a daunting subject, especially when it comes to vectors and calculations. One of the important concepts in physics is displacement, which is defined as the change in position of an object from its initial point to its final point.

If you’re struggling with displacement or just want to brush up on your physics skills, you’ve come to the right place. In this article, we’ll explore the basics of displacement, how to calculate it, and some common examples to help you fully understand this concept. So let’s get started!

## What is Displacement?

Displacement is a vector quantity that measures the change in position of an object. It is different from distance, which is a scalar quantity that measures the total path traveled by an object.

• Displacement is a vector, meaning it has a magnitude and direction
• Displacement is always measured from the starting point to the ending point, regardless of the path taken
• Displacement is usually represented by the symbol Δx

## Calculating Displacement

There are different ways to calculate displacement depending on the given information. The most common methods are using the distance and direction, or using the coordinates of the starting and ending points.

### Method 1: Distance and Direction

If you know the distance and direction of an object’s travel, you can calculate the displacement using trigonometry. Let’s say an object moves 10 meters to the right (positive x-axis) and 5 meters upwards (positive y-axis). We can represent this displacement as:

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Δx = 10 m (to the right)
Δy = 5 m (upwards)

Using the Pythagorean theorem, we can calculate the magnitude of the displacement:

Δs = √(Δx² + Δy²) = √(10² + 5²) = √125

To find the direction, we can use the tangent function:

θ = tan⁻¹(Δy/Δx) = tan⁻¹(5/10) = 26.57° above the x-axis

### Method 2: Coordinates of Starting and Ending Points

If you know the coordinates of the starting point and ending point, you can calculate the displacement using the distance formula:

Δs = √[(x₂ – x₁)² + (y₂ – y₁)²]

Let’s say an object starts at (2, 3) and ends at (-4, 1). We can represent this as:

x₁ = 2, y₁ = 3
x₂ = -4, y₂ = 1

Δs = √[(-4 – 2)² + (1 – 3)²] = √36 + 4 = √40

To find the direction, we can use the inverse tangent function:

θ = tan⁻¹[(y₂ – y₁)/(x₂ – x₁)] = tan⁻¹[(1 – 3)/(-4 – 2)] = tan⁻¹(-0.33) = -19.11° below the x-axis

## Examples of Displacement

Displacement is a common concept in physics and can be observed in various scenarios. Here are some examples:

### Example 1: A Biker’s Displacement

A cyclist travels 5 km towards the east and then 3 km towards the north. What is the cyclist’s displacement?

• Δx = 5 km (to the right)
• Δy = 3 km (upwards)
• Δs = √(5² + 3²) = √34
• θ = tan⁻¹(3/5) = 30.96° above the x-axis

The cyclist’s displacement is √34 km at 30.96° above the x-axis.

### Example 2: A Car’s Displacement

A car starts at point A (2, 3) and moves to point B (-4, 1). What is the car’s displacement?

• x₁ = 2, y₁ = 3
• x₂ = -4, y₂ = 1
• Δs = √[(−4−2)²+(1−3)²] = √40
• θ = tan⁻¹[(1−3)/(−4−2)] = −19.11° below the x-axis

The car’s displacement is √40 units at -19.11° below the x-axis.

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