Firstly before solving this question, teach all the pre requisites ( eg basics of all graphs used in kinematics, tricks to solve graphs very quickly ) and then explain this question in detail. Try to explain everything like Richard feynman. ---**Question Stem:** 26. The graph shows the variation of v² v/s x. The magnitude of maximum acceleration is α m/s². Value of α. **Chart Description:** * **Type:** Line graph. * **Coordinate Axes:** * X-axis represents position, labeled "x(m)". The axis starts at 0 and has marked points at 2, 3, and 5. An arrow indicates the positive direction. * Y-axis represents the square of velocity, labeled "v² (m²/s²)". The axis starts at 0 and has marked points at 4 and 16. An arrow indicates the positive direction. * **Data Series:** The graph consists of three connected line segments: * Segment 1: A horizontal line from (0, 4) to (2, 4). This represents v² = 4 for 0 ≤ x ≤ 2. Dotted lines connect point (2, 4) to the x and y axes. * Segment 2: A straight line from (2, 4) to (3, 16). This represents v² increasing linearly with x from x=2 to x=3. Dotted lines connect point (3, 16) to the x and y axes. * Segment 3: A straight line from (3, 16) to (5, 0). This represents v² decreasing linearly with x from x=3 to x=5, reaching v²=0 at x=5. A dotted line connects point (5, 0) to the x axis (though it lies on it). **Other Relevant Information (from the problem context and graph analysis):** The relationship between acceleration (a), velocity (v), and position (x) is given by the kinematic equation a = v (dv/dx). The graph is of v² versus x. Let y = v². Then y = v² and v = √y. dv/dx = d(√y)/dx = (1 / (2√y)) * (dy/dx) = (1 / (2v)) * (d(v²)/dx). So, a = v * (1 / (2v)) * (d(v²)/dx) = (1/2) * (d(v²)/dx). The acceleration at any point is half the slope of the v² versus x graph at that point. **Calculation of Slopes and Acceleration:** * For 0 ≤ x ≤ 2 (Segment 1): The slope d(v²)/dx is 0. Acceleration a = (1/2) * 0 = 0 m/s². * For 2 ≤ x ≤ 3 (Segment 2): The slope d(v²)/dx = (16 - 4) / (3 - 2) = 12 / 1 = 12 m²/s². Acceleration a = (1/2) * 12 = 6 m/s². * For 3 ≤ x ≤ 5 (Segment 3): The slope d(v²)/dx = (0 - 16) / (5 - 3) = -16 / 2 = -8 m²/s². Acceleration a = (1/2) * (-8) = -4 m/s². **Magnitude of Maximum Acceleration:** The magnitudes of acceleration in the different segments are |0| = 0 m/s², |6| = 6 m/s², and |-4| = 4 m/s². The maximum magnitude of acceleration is 6 m/s². The problem states the magnitude of maximum acceleration is α m/s². **Value of α:** α = 6.

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Welcome to kinematics graphs! To understand motion, we need three fundamental quantities: position, velocity, and acceleration. Position tells us where an object is, velocity tells us how fast it's moving, and acceleration tells us how the velocity is changing. The key insight is that velocity is the rate of change of position with respect to time, and acceleration is the rate of change of velocity with respect to time. On a position versus time graph, the slope at any point gives us the velocity at that moment. Now let's explore a crucial relationship that connects acceleration to position-based graphs. When we don't have time as our variable, we use the chain rule from calculus. Since acceleration equals dv over dt, we can rewrite this as dv over dx times dx over dt. But dx over dt is just velocity! So we get the fundamental relationship: acceleration equals velocity times the slope of the velocity versus position graph. For v-squared versus x graphs, this simplifies even further to acceleration equals half the slope of the v-squared versus x graph. Now let's analyze the actual v-squared versus x graph from our problem. The graph has three distinct linear segments. In the first segment from x equals 0 to 2 meters, we have a horizontal line at v-squared equals 4. Since the slope is zero, the acceleration is also zero. In the second segment from x equals 2 to 3 meters, v-squared increases from 4 to 16, giving us a slope of 12. Using our formula, the acceleration is 6 meters per second squared. Finally, in the third segment from x equals 3 to 5 meters, v-squared decreases from 16 to 0, giving us a slope of negative 8 and an acceleration of negative 4 meters per second squared. Now let's find the maximum acceleration magnitude. We calculated three acceleration values: 0, 6, and negative 4 meters per second squared. Taking the absolute values gives us magnitudes of 0, 6, and 4. Comparing these values, the maximum magnitude is clearly 6 meters per second squared, which occurs in segment 2 where the object is speeding up most rapidly. Therefore, alpha equals 6. This systematic approach of calculating slopes, finding accelerations, taking magnitudes, and comparing values is the key to solving v-squared versus x graph problems efficiently. Let me summarize the key techniques for solving v-squared versus x graph problems quickly. First, always remember the fundamental formula: acceleration equals half the slope of the v-squared versus x graph. For linear segments, horizontal lines mean zero acceleration, positive slopes mean positive acceleration, and negative slopes mean negative acceleration. Calculate slopes using rise over run, then multiply by one-half to get acceleration. For problems asking for maximum magnitude, calculate all accelerations, take absolute values, and find the largest. Using these systematic steps, we found that the maximum acceleration magnitude in our problem is 6 meters per second squared, so alpha equals 6.