In this section, we delve into the graphical analysis of motion, focusing on displacement-time and velocity-time graphs. These graphs provide a visual representation of motion, helping us understand how objects move over time.

## Displacement-Time Graph

The displacement of an object is defined as its distance from the initial point in a specific direction. On a displacement-time graph, displacement is the dependent variable on the y-axis, while time is the independent variable on the x-axis. These graphs are also known as position-time graphs. They can depict three different scenarios:

**1. Stationary Object:** The graph shows a horizontal line, indicating that the object's position does not change over time. The slope is zero, signifying that the object's velocity is zero.

**2. Constant Velocity:** The graph is a straight line with a constant positive slope. This indicates that the object moves at a steady rate in a specific direction.

**3. Constant Acceleration:** The graph shows a curve, where the slope increases with time, indicating that the object's velocity is increasing at a constant rate.

The slope of a displacement-time graph is calculated as follows:

$\text{Slope} = \frac{\Delta d}{\Delta t}$Key takeaways from the displacement-time graph include:

- The slope represents velocity.
- A constant velocity is depicted by a straight line, whereas acceleration is shown by curved lines.
- A positive slope indicates motion in a positive direction, a negative slope indicates motion in a negative direction, and a zero slope indicates that the object is at rest.

Image courtesy of BYJUS

**Example 1: Displacement-Time Graph Problem**

A car starts from rest and moves in a straight line with a constant acceleration of $2 \, \text{m/s}^2$ for $5$ seconds. Plot the displacement-time graph and calculate the total displacement of the car during this period.

**Solution:**

**1. Calculate Total Displacement:**

Use the formula for displacement with constant acceleration, $s = ut + \frac{1}{2}at^2$, where $u = 0 \, \text{m/s}$ (initial velocity), $a = 2 \, \text{m/s}^2$(acceleration), and $t = 5 \, \text{s}$ (time).

$s = 0 \times 5 + \frac{1}{2} \times 2 \times 5^2 = \frac{1}{2} \times 2 \times 25 = 25 \, \text{m}$**2. Plot the Graph:**

The displacement-time graph will start at the origin (since displacement is zero at $t=0$) and curve upwards, reflecting constant acceleration. At $t=5$, the displacement is $25 \, \text{m}$.

## Velocity-Time Graph

In a velocity-time graph, velocity is the dependent variable on the y-axis, and time is the independent variable on the x-axis. The slope of this graph is calculated as:

$\text{Slope} = \frac{\Delta v}{\Delta t}$This slope represents acceleration, leading to the following conclusions:

- A steep slope indicates a rapid change in velocity.
- A shallow slope indicates a slow change in velocity.
- A negative slope indicates that the object is decelerating.
- A positive slope indicates that the object is accelerating.
- The area under the graph represents the total displacement of the object during the given time period.

Image courtesy of BYJUS

### Example 2: Velocity-Time Graph Problem

A cyclist moves with a constant velocity of $10 \, \text{m/s}$ for $4$ seconds, then accelerates at $3 \, \text{m/s}^2$ for the next $2$ seconds. Plot the velocity-time graph and calculate the total displacement during these $6$ seconds.

**Solution:**

**1. Calculate Displacement for Each Phase:**

**Constant Velocity Phase (First (4) seconds):**

Displacement is $s_1 = v \times t = 10 \times 4 = 40 \, \text{m}$.

**Acceleration Phase (Next (2) seconds):**

Use $s_2 = ut + \frac{1}{2}at^2$ where $u = 10 \, \text{m/s}$, $a = 3 \, \text{m/s}^2$, and $t = 2 \, \text{s}$.

$s_2 = 10 \times 2 + \frac{1}{2} \times 3 \times 2^2 = 20 + 6 = 26 \, \text{m}$

**2. Calculate Total Displacement:**

**3. Plot the Graph:**

The velocity-time graph shows a horizontal line at $10 \, \text{m/s}$ for the first $4$seconds, then a straight line sloping upwards for the next $2$ seconds, reflecting the acceleration.

Rahil spent ten years working as private tutor, teaching students for GCSEs, A-Levels, and university admissions. During his PhD he published papers on modelling infectious disease epidemics and was a tutor to undergraduate and masters students for mathematics courses.