Let's analyze the problem step by step: **Given:** - A particle moves along the x-axis. - The particle experiences **variable acceleration** as shown in an **acceleration-time (\(a-t\)) graph**. - You are asked: **What does the area under the graph, between \(t_1\) and \(t_2\), represent?** Let's review the concepts: --- ### **Step 1: Understanding the Area Under a Curve** - In general, the area under a curve in a graph represents the **integral** of the function plotted on the y-axis with respect to the variable on the x-axis. #### **For an \(a-t\) graph:** - y-axis: **acceleration (\(a\))** - x-axis: **time (\(t\))** - The area under the curve from \(t_1\) to \(t_2\) is: \[ \int_{t_1}^{t_2} a\,dt \] --- ### **Step 2: Mathematical Meaning** - By definition, the **integral of acceleration with respect to time** gives the **change in velocity**. - Mathematically: \[ \int_{t_1}^{t_2} a\,dt = v_2 - v_1 \] where \(v_1\) is the velocity at \(t_1\), and \(v_2\) is the velocity at \(t_2\). --- ### **Step 3: Matching with the Options** Let's check each option: **A) average acceleration** - No, the average is the mean value, not the area. **B) \(v_2-v_1\)** - Yes, this matches our finding! **C) \(x_2-x_1\)** - No, this would be the area under a velocity-time graph. **D) \(a_2-a_1\)** - No, this is the change in acceleration, not represented by the area. **E) average velocity** - No, again, not represented by the area here. --- ### **Final Answer** \[ \boxed{B)~v_2 - v_1} \] **Explanation:** The area under the acceleration-time graph between \(t_1\) and \(t_2\) gives the change in velocity (\(v_2-v_1\)) of the particle during that time interval.