Displacement-time graphs

Displacement-time graphs are used to represent the motion of an object over time. The horizontal axis (x-axis) represents time, while the vertical axis (y-axis) represents displacement, which is the distance from a reference point in a specific direction. These graphs are useful for understanding how an object moves and for calculating its velocity and speed.

1. Understanding Displacement-Time Graphs:

In a displacement-time graph, the slope of the graph indicates the velocity of the object. The displacement is measured relative to a starting point (often called the origin). A straight line in the graph indicates constant motion, while a curved line indicates changing motion (acceleration or deceleration).

2. Interpreting the Graph:

• If the graph is a straight line, the object is moving at a constant velocity.
• If the graph is curved, the object's velocity is changing, meaning it is accelerating or decelerating.
• If the line is horizontal, the object is stationary (no displacement). This means the velocity is zero.
• If the line is sloping upwards, the object is moving in a positive direction. If the line slopes downwards, the object is moving in the opposite direction (negative displacement).

3. Finding the Velocity from a Displacement-Time Graph:

The velocity of an object is the rate of change of displacement with respect to time. In terms of a graph, the velocity is found by calculating the slope of the displacement-time graph. The formula for velocity is:

\[ \text{Velocity} = \frac{\text{Change in Displacement}}{\text{Change in Time}} = \frac{\Delta y}{\Delta x} \]

Where \( \Delta y \) is the change in displacement (vertical change) and \( \Delta x \) is the change in time (horizontal change).

4. Example: Finding Velocity:

Consider the following displacement-time graph where the object moves in a straight line, starting at a displacement of 0 m at \( t = 0 \) and reaching a displacement of 20 m at \( t = 5 \) seconds. The line is straight, indicating constant velocity.

Step 1: Calculate the change in displacement:

Change in displacement \( \Delta y = 20 \, \text{m} - 0 \, \text{m} = 20 \, \text{m} \).

Step 2: Calculate the change in time:

Change in time \( \Delta x = 5 \, \text{seconds} - 0 \, \text{seconds} = 5 \, \text{seconds} \).

Step 3: Use the formula for velocity:

Velocity \( = \frac{\Delta y}{\Delta x} = \frac{20 \, \text{m}}{5 \, \text{seconds}} = 4 \, \text{m/s} \).

Therefore, the velocity of the object is \( 4 \, \text{m/s} \) in the positive direction.

5. Finding the Speed from a Displacement-Time Graph:

Speed is the scalar quantity that represents the magnitude of velocity. Speed is always positive and does not have a direction, whereas velocity is a vector quantity and has direction.

If the object is moving at a constant speed, the magnitude of the velocity is the same as the speed. Therefore, the speed can be calculated in the same way as velocity by determining the slope of the graph. If the object changes direction, the speed is still the magnitude of the rate of change of displacement, which is the slope of the graph regardless of the direction of motion.

6. Example: Finding Speed:

Let’s say the displacement-time graph is for an object that moves from \( 0 \, \text{m} \) at \( t = 0 \) to \( 20 \, \text{m} \) at \( t = 5 \) seconds, and then moves from \( 20 \, \text{m} \) back to \( 0 \, \text{m} \) at \( t = 10 \) seconds (indicating a return journey). The graph forms a straight line from \( t = 0 \) to \( t = 5 \), and another straight line from \( t = 5 \) to \( t = 10 \). The speed is calculated as the slope of the graph, which is the same as velocity when the object moves at a constant rate in each section.

Step 1: Find the speed for the first part of the journey (from \( 0 \, \text{m} \) to \( 20 \, \text{m} \)):

The change in displacement is \( 20 \, \text{m} - 0 \, \text{m} = 20 \, \text{m} \), and the change in time is \( 5 \, \text{seconds} - 0 \, \text{seconds} = 5 \, \text{seconds} \).

The speed is: \[ \text{Speed} = \frac{\Delta y}{\Delta x} = \frac{20 \, \text{m}}{5 \, \text{seconds}} = 4 \, \text{m/s} \]

Step 2: Find the speed for the return journey (from \( 20 \, \text{m} \) back to \( 0 \, \text{m} \)):

The change in displacement is \( 0 \, \text{m} - 20 \, \text{m} = -20 \, \text{m} \), and the change in time is \( 10 \, \text{seconds} - 5 \, \text{seconds} = 5 \, \text{seconds} \).

The speed is the same magnitude: \[ \text{Speed} = \frac{20 \, \text{m}}{5 \, \text{seconds}} = 4 \, \text{m/s} \]

Note that the speed is always positive, regardless of the direction of travel.

7. Special cases to note:

• If the graph is a straight line, the velocity and speed are constant and can be easily calculated from the slope.
• If the graph is curved, the velocity is changing, and you will need to calculate the instantaneous velocity at any given point by finding the slope of the tangent to the curve at that point.
• If the object moves in a direction and then returns, the displacement at the end of the journey may be zero, but the total distance travelled is the sum of all movements along the path.