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 $4.333\text{sec}$. The total distance covered by the body in $6\text{s}$ is

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

Answer 1

The correct option is A $\frac{37}{3}\text{m}$

As we know that total distance covered by the body is given by,

$s=\int |V|dt=$ sum of magnitude of area under $v-t$ graph

$s=|\text{area of OABSO}|+|\text{area of SCDS}$

$s=\frac{1}{2}\left(4\right)(4.333+1)+\frac{1}{2}(6-4.333)\left(2\right)$

$\therefore s=12.333=\frac{37}{3}\text{m}$

Hence, option $\left(\text{A}\right)$ is correct.