Question:

From the v - t graph shown, the ratio of distance to displacement in 25 s of motion is:
 ratio of distance to displacement in 25 s of motion

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When finding distance, always consider the absolute value of the displacement, i.e., sum the areas without sign.
Updated On: Mar 22, 2025
  • \( \frac{3}{5} \)
  • \( \frac{1}{2} \)
  • \( \frac{2}{5} \)
  • \( \frac{1}{3} \)
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The Correct Option is C

Solution and Explanation

Displacement is the area under the graph and distance is the total area covered, including all positive and negative regions of the graph.
Displacement = Area of graph with sign \[ = \left(\frac{1}{2} \times 10 \times 5 \right) + \left(10 \times 5 \right) = 25 + 50 = 75 \, \text{m} \] Distance = Area of graph with positive value \[ = \left(\frac{1}{2} \times 5 \times 20 \right) + \left(5 \times 20 \right) = 25 + 50 = 250 \, \text{m} \] Now, the ratio of distance to displacement is: \[ \frac{250}{75} = \frac{2}{5} \] Thus, the ratio is \( \frac{2}{5} \).
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