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JEE Mains · Physics · STD 11 - Motion in straight line
  • JEE Mains 2023
  • Medium
  • MCQ

From the \(v\) - \(t\) graph shown. the ratio of distance to displacement in \(25\,s\) of motion

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Answer & Solution

Correct Answer

(C) \(\frac{5}{3}\)

Step-by-step Solution

Detailed explanation

Area under the graph from \(t=0\) to \(t=20\,sec =200\,m\)

Area under the graph from \(t=20\) to \(t=25\,sec =50\,m\)

So distance covered \(=(200+50) m =250\,m\)

Displacement \(=(200-50) m =150\,m\)

\(\frac{250}{150}=\frac{5}{3}\)

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