Distance from Velocity-Time Graph | Motion in 1D | JEE Physics | Doubtify JEE

 

📈 Velocity-Time Graph Problem | Motion in One Dimension | Physics | Doubtify JEE

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💡 Question:

The velocity (v) and time (t) graph of a body in a straight-line motion is shown in the figure. Point S is at 4.333 sec. The total distance covered by the body in 6 seconds is?

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🖼️ Question Image:

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🧠 Solution Image:

✍️ Detailed Explanation:

To find the total distance, we analyze the area under the velocity-time graph. Remember, the area above the time axis (positive velocity) adds to distance, and the area below (negative velocity) also adds to distance—just as a positive value.

Step-by-step:

1. From 0 to 4.333 s (point S):

   • Velocity is positive.

   • Use the area of a triangle or trapezium, depending on the graph.

2. From 4.333 to 6 s:

   • If velocity is negative here, it implies a reversal in direction.

   • We still take the absolute value of the area for distance.

3. Total Distance = Area(0 to 4.333 s) + |Area(4.333 to 6 s)|

Use appropriate area formulas:

• Triangle: $\frac{1}{2} \times \text{base} \times \text{height}$

• Trapezium: $\frac{1}{2} \times (\text{sum of parallel sides}) \times \text{height}$

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🧠 Pro Tip:

Understanding how to interpret velocity-time graphs not only helps in JEE but builds your intuition for real-world motion problems. Always look for turning points (like where velocity becomes zero) as these can mark direction changes.

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🎥 Video Solution:

🔍 Why this Question is Important:

• Tests core concepts of kinematics and graphical analysis.

• Helps students visualize motion instead of memorizing formulas.

• Common in JEE Mains and school-level Physics exams.

• Distinguishes between distance and displacement.

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