JEE Mains · Physics · STD 11 - 2. motion in straight line
From the \(v\) - \(t\) graph shown. the ratio of distance to displacement in \(25\,s\) of motion
- A \(\frac{3}{5}\)
- B \(\frac{1}{2}\)
- C \(\frac{5}{3}\)
- D \(1\)
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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