MHT CET · Physics · Motion In One Dimension
For the velocity-time graph shown in the figure below, the distance covered by the body in last two second of its motion is ' \(\mathrm{S}_1\) '. What is the ratio of ' \(S_1\) ' to the total distance covered by it
- A \(\frac{1}{2}\)
- B \(\frac{1}{4}\)
- C \(\frac{1}{3}\)
- D \(\frac{2}{3}\)
Answer & Solution
Correct Answer
(B) \(\frac{1}{4}\)
Step-by-step Solution
Detailed explanation
Distance covered = Area enclosed by velocity-time graph.
Total distance covered in 6 seconds,
\(S=\frac{1}{2} \times 2 \times 10+2 \times 10+\frac{1}{2} \times 2 \times 10=40 m\)
Distance covered in last two seconds,
\(\begin{aligned}
& S^{\prime}=\frac{1}{2} \times 2 \times 10=10 \mathrm{~m} \\
& \therefore \text { Fraction of distance }=\frac{S^{\prime}}{S}=\frac{10}{40}=\frac{1}{4}
\end{aligned}\)
Total distance covered in 6 seconds,
\(S=\frac{1}{2} \times 2 \times 10+2 \times 10+\frac{1}{2} \times 2 \times 10=40 m\)
Distance covered in last two seconds,
\(\begin{aligned}
& S^{\prime}=\frac{1}{2} \times 2 \times 10=10 \mathrm{~m} \\
& \therefore \text { Fraction of distance }=\frac{S^{\prime}}{S}=\frac{10}{40}=\frac{1}{4}
\end{aligned}\)
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