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To determine the average force acting and displacement of the object between \( t = 2 \, \text{s} \) and \( t = 4 \, \text{s} \), let’s proceed with the following steps:

### 1. Analyze Velocity Change to Find Acceleration
From the velocity-time graph:
- At \( t = 2 \, \text{s} \), the velocity \( v_1 = 8 \, \text{m/s} \).
- At \( t = 4 \, \text{s} \), the velocity \( v_2 = 4 \, \text{m/s} \).

The change in velocity (\( \Delta v \)) over this time interval is:
\[
\Delta v = v_2 - v_1 = 4 - 8 = -4 \, \text{m/s}
\]

The time interval (\( \Delta t \)) between \( t = 2 \) and \( t = 4 \) seconds is \( \Delta t = 4 - 2 = 2 \, \text{s} \).

The acceleration \( a \) is then calculated as:
\[
a = \frac{\Delta v}{\Delta t} = \frac{-4}{2} = -2 \, \text{m/s}^2
\]

### 2. Calculate Average Force Using Newton's Second Law
Using the mass \( m = 2.5 \, \text{kg} \) and the calculated acceleration:
\[
F = m \cdot a = 2.5 \times (-2) = -5 \, \text{N}
\]

So, the average force acting on the object between \( t = 2 \, \text{s} \) and \( t = 4 \, \text{s} \) is \( -5 \, \text{N} \).

### 3. Calculate Displacement Using the Area Under the Graph
The displacement between \( t = 2 \, \text{s} \) and \( t = 4 \, \text{s} \) can be found by calculating the area under the velocity-time graph over this interval.

Between \( t = 2 \, \text{s} \) and \( t = 4 \, \text{s} \):
- The shape under the graph is a trapezoid with one side at \( v = 8 \, \text{m/s} \), the other at \( v = 4 \, \text{m/s} \), and a height (duration) of \( 2 \, \text{s} \).

The area \( A \) of a trapezoid is:
\[
A = \frac{1}{2} \times ( \text{Base}_1 + \text{Base}_2 ) \times \text{Height}
\]
\[
A = \frac{1}{2} \times (8 + 4) \times 2 = \frac{1}{2} \times 12 \times 2 = 12 \, \text{m}
\]

Thus, the displacement between \( t = 2 \, \text{s} \) and \( t = 4 \, \text{s} \) is \( 12 \, \text{m} \).

### Summary of Results
- **Average Force**: \( -5 \, \text{N} \)
- **Displacement**: \( 12 \, \text{m} \)

Would you like further details on these calculations or have any questions?

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Here are some related questions to explore further:
1. What is the total displacement of the object from \( t = 0 \) to \( t = 6 \, \text{s} \)?
2. How would the result change if the mass of the object were different?
3. What is the significance of the negative sign in the calculated force?
4. How can you determine the total distance traveled from a velocity-time graph?
5. How would you calculate the work done by this force over the interval \( t = 2 \, \text{s} \) to \( t = 4 \, \text{s} \)?

**Tip:** To find the displacement on a velocity-time graph, always calculate the area under the curve within the given time interval.

Velocity-time graph of an object of mass 2.5 kg moving in a straight line is shown in the figure. The average force acting and displacement of the object between t = 2 s to t = 4 s is respectively.

Learn how to calculate the average force and displacement of a 2.5 kg object from a velocity-time graph between t = 2 s and t = 4 s. This problem explores key kinematic concepts, Newton’s laws, and graph analysis. Ideal for Grades 10-12.

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