JEE Mains · Physics · STD 11 - 2. motion in straight line
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\(.......m\)
- A \(12\)
- B \(\frac{49}{4} m\)
- C \(11\)
- D \(\frac{37}{3} m\)
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 \(=\left| A _{1}\right|+\left| A _{2}\right|\) \(A _{1}=\frac{1}{2}\left[\frac{13}{3}+1\right] 4=\frac{32}{3}\)…
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