For the velocity-time graph shown in the figure below, the distance covered by the body in last two second…
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
$\frac{1}{2}$
$\frac{1}{4}$
$\frac{1}{3}$
$\frac{2}{3}$
Solution
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,
$\begin{aligned}
& S^{\prime}=\frac{1}{2} \times 2 \times 10=10 \mathrm{~m} \\
& \therefore \text { Fraction of distance }=\frac{S^{\prime}}{S}=\frac{10}{40}=\frac{1}{4}
\end{aligned}$
