Question:
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 $6 \mathrm{~s}$ is:
Correct Option: 1
Solution:
$O S=4+\frac{1}{3}=\frac{13}{3}$
$S D=2-\frac{1}{3}=\frac{5}{3}$
Distance covered by the body $=$ area of $v-t$ graph
$=\operatorname{ar}(O A B S)+\operatorname{ar}(S C D)$
$=\frac{1}{2}\left(\frac{13}{3}+1\right) \times 4+\frac{1}{2} \times \frac{5}{3} \times 2$
$=\frac{32}{3}+\frac{5}{3}=\frac{37}{3} \mathrm{~m}$