From the above graph, what is the distance covered in the last stage of the motion?
Question 1 Report
From the above graph, what is the distance covered in the last stage of the motion?
Answer Details
The graph is a velocity-time graph with three stages of motion. From 0 to 10 seconds the body moves at a constant velocity of 12 m/s. From 10 to 14 seconds the velocity drops linearly from 12 m/s to 0 m/s, representing uniform deceleration. This final segment is the last stage of the motion.
The distance covered during any stage equals the area under the velocity-time curve for that interval. The last stage forms a right triangle with base \(\Delta t = 14 - 10 = 4\) s and height \(v = 12\) m/s.
Distance = \(\frac{1}{2} \times 4 \times 12 = 24\) m.
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