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JEE Main 2026 (5 Apr, Shift 2) · Graphs of Motion

Graphs of Motion question from JEE Main 2026 (5 Apr, Shift 2)

Q2JEE Main 2026 · 5 APR · SHIFT 2Easy

Graphs of Motion

The velocity (v)(v) versus time (t)(t) plot of a particle is shown in the figure, for a time interval of 40 s. The total distance travelled by the particle and the average velocity during this period are, respectively

______.

Figure

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Answer: (c)

  1. The graph has four triangles, each of base 10 s10\text{ s} and height 5 m/s5\text{ m/s}: area =12×10×5=25 m= \frac{1}{2}\times 10\times 5 = 25\text{ m} each.
  2. Distance = sum of magnitudes =4×25=100 m= 4\times 25 = 100\text{ m}
  3. Displacement = 25+25−25−25=025 + 25 - 25 - 25 = 0, so average velocity =0= 0
  4. So the answer is (c) 100 m and zero.
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