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Displacement

CSCA Displacement study guide organized around the publicly available CSCA syllabus. Practice Physics questions on aicsca.com.

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Displacement

1. Core Concept: Displacement vs. Distance

This is the most fundamental yet easily confused concept in kinematics. You must distinguish between "change in position" and "actual path traveled."

| Aspect | Displacement ($\vec{s}$ or $\vec{x}$) | Distance (Path Length, $d$) |

| :--- | :--- | :--- |

| **Definition** | Directed line segment from **Initial** to **Final** position | The total length of the actual **trajectory** |

| **Nature** | **Vector**: Has both magnitude and direction | **Scalar**: Has magnitude only, no direction |

| **Sign (+/-)** | Indicates Direction (in 1D motion) | Always Positive (Non-negative) |

| **Dependence** | Depends only on Start and End points | Depends on the specific path taken |

| **Relationship** | $|\text{Displacement}| \le \text{Distance}$ | Equal only in **Single-Direction Linear Motion** |

**⚠️ CSCA Exam Trap**:

If the motion involves a "U-turn" or "reversal" (e.g., vertical throw turning back down, or a car braking and reversing), **Distance is strictly greater than Displacement magnitude**. You must calculate distance in segments, whereas displacement only cares about the start and end.

2. Representation and Calculation

#### A. One-Dimensional (1D) Motion

In a linear coordinate system, displacement is calculated by the difference in coordinates. **The sign (+/-) indicates direction.**

$$ \Delta x = x_{\text{final}} - x_{\text{initial}} $$

* $\Delta x > 0$: Displacement is in the positive direction.

* $\Delta x < 0$: Displacement is in the negative direction.

#### B. Two-Dimensional (2D) Motion

In 2D, displacement is the result of vector addition. We typically use the Pythagorean theorem and trigonometry.

Given Initial Position $A(x_1, y_1)$ and Final Position $B(x_2, y_2)$:

* **Vector Form**: $\vec{s} = (x_2 - x_1)\hat{i} + (y_2 - y_1)\hat{j}$

* **Magnitude**: $$ |\vec{s}| = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2} $$

* **Direction**: If $\theta$ is the angle with the x-axis, then $\tan \theta = \frac{\Delta y}{\Delta x}$ (check quadrants).

3. Typical Examples

**Example 1: The "Turning Point" Trap in 1D**

A particle moves along the x-axis with position function $x(t) = 2t^2 - 4t + 1$ (m). Find (1) Displacement and (2) Distance from $t=0$ to $t=3\text{s}$.

**Strategy**:

1. **Displacement**: Just check endpoints.

2. **Distance**: Find the "Turning Point" (where velocity $v=0$).

**Solution**:

(1) **Displacement**:

$$x(0) = 1\text{m}, \quad x(3) = 2(3)^2 - 4(3) + 1 = 7\text{m}$$

$$\Delta x = x(3) - x(0) = 7 - 1 = +6\text{m} \quad (\text{Direction: +x})$$

(2) **Distance**:

Find velocity: $v(t) = x'(t) = 4t - 4$.

Set $v(t) = 0 \Rightarrow t=1\text{s}$. The particle **turns around** at $t=1\text{s}$.

* **Segment 1 ($0-1\text{s}$)**: $x(1) = -1\text{m}$. Displacement $\Delta x_1 = -1 - 1 = -2\text{m}$. Distance $d_1 = |-2| = 2\text{m}$.

* **Segment 2 ($1-3\text{s}$)**: $x(3) = 7\text{m}$. Displacement $\Delta x_2 = 7 - (-1) = 8\text{m}$. Distance $d_2 = |8| = 8\text{m}$.

* **Total Distance**: $d = d_1 + d_2 = 2 + 8 = 10\text{m}$.

**Example 2: 2D Vector Addition**

A person walks $40\text{m}$ East, then $30\text{m}$ North. Find total displacement and distance.

**Solution**:

* **Distance (Scalar)**: $40 + 30 = 70\text{m}$.

* **Displacement (Vector)**: Hypotenuse of the right triangle.

* Magnitude: $s = \sqrt{40^2 + 30^2} = 50\text{m}$.

* Direction: $\tan \theta = \frac{30}{40} = 0.75 \Rightarrow \theta \approx 37^\circ$ (North of East).