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MHT CET 2024 (11 May, Shift 1) · Graphs of Motion

Graphs of Motion question from MHT CET 2024 (11 May, Shift 1)

Q11MHT CET 2024 · 11 MAY · SHIFT 1Easy

Graphs of Motion

For the velocity-time graph shown in the figure below, the distance covered by the body in last two second of its motion is ' S1S_1 '. What is the ratio of ' S1S_1 ' to the total distance covered by it

Figure

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Answer: (b)

  1. Distance = area under the vv-tt graph. Read the time axis as A at 1 s1\text{ s}, B at 3 s3\text{ s}, C at 5 s5\text{ s}, D at 7 s7\text{ s} (the last label '4' is a misprint for 7).
  2. Total distance = trapezium area =12(2+6)×10=40 m= \frac{1}{2}(2 + 6)\times 10 = 40\text{ m}
  3. Last two seconds (5 s to 7 s): S1=12×2×10=10 mS_1 = \frac{1}{2}\times 2\times 10 = 10\text{ m}
  4. S1S=1040=14\frac{S_1}{S} = \frac{10}{40} = \frac{1}{4}. So the answer is (b).
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