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JEE Mains · Physics · STD 11 - Motion in straight line
  • JEE Mains 2020
  • Medium
  • MCQ

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\(.......m\)

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Answer & Solution

Correct Answer

(D) \(\frac{37}{3} m\)

Step-by-step Solution

Detailed explanation

\(OS =4+\frac{1}{3}=\frac{13}{3}\)

\(SD =2-\frac{1}{3}=\frac{5}{3}\)

Area of \(OABS\) is \(A _{1}\)

Area of \(SCD\) is \(A _{2}\)

Distance \(=\left| A _{1}\right|+\left| A _{2}\right|\)

\(A _{1}=\frac{1}{2}\left[\frac{13}{3}+1\right] 4=\frac{32}{3}\)

\(A _{2}=\frac{1}{2} \times \frac{5}{3} \times 2=\frac{5}{3}\)

Distance \(=\left|A_{1}\right|+\left|A_{2}\right|\)

\(=\frac{32}{3}+\frac{5}{3}\)

\(=\frac{37}{3}\)

 

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