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2.Motion in Straight Line
medium
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
$\frac{37}{3} m$
(JEE MAIN-2020)
Solution
$OS =4+\frac{1}{3}=\frac{13}{3}$
$SD =2-\frac{1}{3}=\frac{5}{3}$
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 $=\left|A_{1}\right|+\left|A_{2}\right|$
$=\frac{32}{3}+\frac{5}{3}$
$=\frac{37}{3}$
Standard 11
Physics
