Velocity-Time Graphs

Understanding Velocity-Time Graphs

A velocity-time graph shows how an object's velocity changes over time. The horizontal axis (x-axis) represents time in seconds (s), while the vertical axis (y-axis) represents velocity in metres per second (m/s).

Velocity can be positive or negative:

• Positive velocity means the object is moving in the forward direction (e.g., a car moving forwards).
• Negative velocity means the object is moving in the opposite direction (e.g., a car reversing).

The shape of the graph tells us about the motion:

• A flat horizontal line means the velocity is constant (no change over time).
• A sloped straight line means the velocity is changing at a constant rate (constant acceleration or deceleration).
• A curved line means the velocity is changing at a varying rate (non-uniform acceleration).

For example, if a car’s velocity-time graph is a flat line at 20 m/s, it means the car is moving forwards at a steady 20 m/s.

Interpreting Graph Features

Gradient of the graph represents acceleration.

Acceleration is a vector quantity that describes the rate of change of velocity with time. It is positive if the velocity increases in the positive direction and negative if the velocity decreases or increases in the negative direction (deceleration).

Mathematically, the gradient is:

$$\text{Acceleration} = \frac{\text{Change in velocity}}{\text{Change in time}} = \frac{\Delta v}{\Delta t}$$

Area under the graph represents displacement.

Displacement is the overall change in position and can be found by calculating the area between the velocity-time graph and the time axis.

If the velocity is positive, the area is above the time axis and displacement is positive; if velocity is negative, the area is below the time axis and displacement is negative.

The zero velocity line (the x-axis) means the object is momentarily stationary — it has zero velocity at that instant.

For instance, if a velocity-time graph shows a straight line sloping upwards from 0 m/s at 0 seconds to 10 m/s at 5 seconds, the acceleration is:

$$\text{Acceleration} = \frac{10 - 0}{5 - 0} = 2 \text{ m/s}^2$$

Types of Motion from Graphs

You can describe different types of motion by looking at the shape and position of the velocity-time graph:

• Constant velocity: The graph is a flat horizontal line, showing velocity does not change over time.
• Acceleration: The graph has a positive gradient (line slopes upwards), meaning velocity is increasing.
• Deceleration: The graph has a negative gradient (line slopes downwards), meaning velocity is decreasing.

For example, a car slowing down will have a velocity-time graph with a line sloping downwards towards zero velocity.

Calculating from Graphs

Finding acceleration from the gradient:

To calculate acceleration, pick two points on the graph and use:

$$\text{Acceleration} = \frac{\text{Change in velocity}}{\text{Change in time}} = \frac{v_2 - v_1}{t_2 - t_1}$$

Note that the sign of acceleration depends on the direction of velocity change: a positive gradient means velocity is increasing in the positive direction, while a negative gradient means velocity is decreasing or increasing in the negative direction.

Calculating displacement from the area under the graph:

The area under the velocity-time graph can be found by splitting the shape into simple shapes like rectangles, triangles, or trapeziums, then calculating their areas and adding or subtracting as needed.

The total area gives the displacement during the time interval.

Using the graph to describe motion:

By combining the gradient and area information, you can describe how an object moves, whether it speeds up, slows down, or moves backwards, and how far it travels from its starting point.

For example, if a velocity-time graph shows a line sloping down from 20 m/s to 0 m/s over 4 seconds, the acceleration is:

$$\text{Acceleration} = \frac{0 - 20}{4 - 0} = -5 \text{ m/s}^2$$

This is a deceleration of 5 m/s².

The displacement is the area under the graph, which is a triangle with base 4 s and height 20 m/s:

$$\text{Displacement} = \frac{1}{2} \times 4 \times 20 = 40 \text{ m}$$