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 $6s$ is :

• A. $12m$
• B. $\frac{49}{4}m$
• C. $11m$
• D. $\frac{37}{3}m$

Answer 1

Answer: (D)

$OS=4+\frac{1}{3}=\frac{13}{3}$
$SD=2-\frac{1}{3}=\frac{5}{3}$
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 $=\left|A_1\right|+\left|A_2\right|$
$=\frac{32}{3}+\frac{5}{3}=\frac{37}{3}$