Worked example 2.1: Velocity-time graph - Free Printable

Problem Analysis:

The image provided is a velocity-time graph, where the vertical axis represents velocity in meters per second ($v$ in m/s) and the horizontal axis represents time in seconds ($t$ in s). The task is to analyze the graph and solve any related problems. Since no specific problem is stated in the question, I will infer common tasks associated with velocity-time graphs, such as:

1. Calculating the total distance traveled.
2. Determining the acceleration during different intervals.
3. Identifying the time intervals of constant velocity.

Step-by-Step Solution:

#### 1. Understanding the Graph:
The graph shows the velocity of an object as a function of time. Key observations from the graph are:
- From \( t = 0 \) to \( t = 2 \) seconds: The velocity increases linearly from 0 m/s to 8 m/s. This indicates a period of constant acceleration.
- From \( t = 2 \) to \( t = 10 \) seconds: The velocity remains constant at 8 m/s. This indicates zero acceleration (constant velocity).
- From \( t = 10 \) to \( t = 12 \) seconds: The velocity decreases linearly from 8 m/s to 4 m/s. This indicates a period of constant deceleration.
- From \( t = 12 \) to \( t = 16 \) seconds: The velocity remains constant at 4 m/s. This indicates zero acceleration (constant velocity).

#### 2. Calculating the Total Distance Traveled:
The total distance traveled by the object can be found by calculating the area under the velocity-time graph. The graph can be divided into geometric shapes for easier calculation:
- From \( t = 0 \) to \( t = 2 \) seconds: The shape is a right triangle.
- Base = 2 seconds
- Height = 8 m/s
- Area = \( \frac{1}{2} \times \text{base} \times \text{height} = \frac{1}{2} \times 2 \times 8 = 8 \) meters

- From \( t = 2 \) to \( t = 10 \) seconds: The shape is a rectangle.
- Width = 8 seconds
- Height = 8 m/s
- Area = \( \text{width} \times \text{height} = 8 \times 8 = 64 \) meters

- From \( t = 10 \) to \( t = 12 \) seconds: The shape is a trapezoid.
- Parallel sides = 8 m/s and 4 m/s
- Height = 2 seconds
- Area = \( \frac{1}{2} \times (\text{sum of parallel sides}) \times \text{height} = \frac{1}{2} \times (8 + 4) \times 2 = \frac{1}{2} \times 12 \times 2 = 12 \) meters

- From \( t = 12 \) to \( t = 16 \) seconds: The shape is a rectangle.
- Width = 4 seconds
- Height = 4 m/s
- Area = \( \text{width} \times \text{height} = 4 \times 4 = 16 \) meters

Adding these areas together gives the total distance:
\[
\text{Total distance} = 8 + 64 + 12 + 16 = 100 \text{ meters}
\]

#### 3. Determining the Acceleration During Different Intervals:
- From \( t = 0 \) to \( t = 2 \) seconds: The acceleration is constant and can be calculated using the formula \( a = \frac{\Delta v}{\Delta t} \).
- Initial velocity (\( v_i \)) = 0 m/s
- Final velocity (\( v_f \)) = 8 m/s
- Time interval (\( \Delta t \)) = 2 seconds
- Acceleration = \( \frac{8 - 0}{2} = 4 \) m/s²

- From \( t = 2 \) to \( t = 10 \) seconds: The velocity is constant, so the acceleration is 0 m/s².

- From \( t = 10 \) to \( t = 12 \) seconds: The deceleration is constant and can be calculated using the formula \( a = \frac{\Delta v}{\Delta t} \).
- Initial velocity (\( v_i \)) = 8 m/s
- Final velocity (\( v_f \)) = 4 m/s
- Time interval (\( \Delta t \)) = 2 seconds
- Acceleration = \( \frac{4 - 8}{2} = -2 \) m/s²

- From \( t = 12 \) to \( t = 16 \) seconds: The velocity is constant, so the acceleration is 0 m/s².

#### 4. Identifying the Time Intervals of Constant Velocity:
- The velocity is constant during the intervals:
- From \( t = 2 \) to \( t = 10 \) seconds
- From \( t = 12 \) to \( t = 16 \) seconds

Final Answer:

The total distance traveled by the object is \(\boxed{100}\) meters.