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JEE Main 2020 (5 Sep, Shift 2) · Graphs of Motion

Graphs of Motion question from JEE Main 2020 (5 Sep, Shift 2)

Q28JEE Main 2020 · 5 SEP · SHIFT 2Medium

Graphs of Motion

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

Figure

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Answer: (d)

  1. Distance = sum of magnitudes of areas under the vv-tt graph.
  2. 0-2 s: 12×2×4=4\frac{1}{2}\times 2\times 4 = 4; 2-3 s: 1×4=41\times 4 = 4; 3 s to 133\frac{13}{3} s: 12×43×4=83\frac{1}{2}\times\frac{4}{3}\times 4 = \frac{8}{3}
  3. 133\frac{13}{3} s to 6 s (below axis): 12×53×2=53\frac{1}{2}\times\frac{5}{3}\times 2 = \frac{5}{3}
  4. Total =8+83+53=373 m= 8 + \frac{8}{3} + \frac{5}{3} = \frac{37}{3}\text{ m}. So the answer is (d).
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