Answer :
Explanation:
To analyze the graph and answer the questions, we'll break down each part step by step. Since I can't directly see the graph, I'll outline how you would approach solving these types of problems generally:
### a) Determine the distance traveled by the object between \( t = 2s \) and \( t = 10s \).
To find the distance traveled, we need to calculate the area under the velocity-time graph between \( t = 2s \) and \( t = 10s \).
1. Identify the velocity values at key points (t = 2s, t = 10s) and the shape of the graph between these times.
2. Calculate the area under the curve using geometric shapes (rectangles, triangles, trapezoids).
Assuming the graph is a combination of straight-line segments:
- If the graph is a straight line or consists of simple shapes, calculate the area of each shape separately and sum them up.
### b) Find the maximum retardation of the object.
Retardation is the negative acceleration. To find the maximum retardation:
1. Identify the points where the slope of the velocity-time graph is negative.
2. Calculate the slope (change in velocity/change in time) in these sections to find the retardation.
3. The steepest negative slope represents the maximum retardation.
### c) Calculate the acceleration of the object.
Acceleration is the slope of the velocity-time graph. To find the acceleration:
1. Identify the segments of the graph where the slope (acceleration) needs to be calculated.
2. For each segment, use the formula:
\[ \text{Acceleration} = \frac{\Delta v}{\Delta t} \]
where \( \Delta v \) is the change in velocity and \( \Delta t \) is the change in time.
### Example Calculation:
Let's assume a hypothetical velocity-time graph for illustration:
1. **From \( t = 2s \) to \( t = 4s \)**:
- Velocity increases linearly from 0 m/s to 4 m/s.
- Area: Triangle with base = 2s, height = 4 m/s.
- Area = \( \frac{1}{2} \times 2 \times 4 = 4 \, \text{m} \).
2. **From \( t = 4s \) to \( t = 8s \)**:
- Velocity is constant at 4 m/s.
- Area: Rectangle with base = 4s, height = 4 m/s.
- Area = \( 4 \times 4 = 16 \, \text{m} \).
3. **From \( t = 8s \) to \( t = 10s \)**:
- Velocity decreases linearly from 4 m/s to 0 m/s.
- Area: Triangle with base = 2s, height = 4 m/s.
- Area = \( \frac{1}{2} \times 2 \times 4 = 4 \, \text{m} \).
Total distance traveled from \( t = 2s \) to \( t = 10s \):
\[ 4 + 16 + 4 = 24 \, \text{m} \]
### Maximum Retardation:
- Assume from \( t = 8s \) to \( t = 10s \), velocity decreases from 4 m/s to 0 m/s.
- Retardation = \(\frac{0 - 4}{10 - 8} = -2 \, \text{m/s}^2 \).
### Acceleration:
- From \( t = 2s \) to \( t = 4s \):
\[ \text{Acceleration} = \frac{4 - 0}{4 - 2} = 2 \, \text{m/s}^2 \].
This is a simplified example. For exact values, refer directly to the given graph and apply these methods accordingly.
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