Question

The velocity $\left(v\right)$ and time $\left(t\right)$ graph of a body in a straight line motion is shown in the figure. The point $S$ is at $4.333\text{s}$ seconds. The total distance covered by the body in $6\text{s}$ is:

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

Answer 1

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

Given
$OS=4.33=\frac{13}{3}\text{s}$
Hence
$SD=2-\frac{1}{3}=\frac{5}{3}\text{s}$
Distance covered bt the body = area of $v-t$ graph = area of $\left(OABS\right)$ + area of $\left(SCD\right)$
$=\frac{1}{2}(\frac{13}{3}+1)\times 4+\frac{1}{2}\times \frac{5}{3}\times 2$
$=\frac{32}{3}+\frac{5}{3}=\frac{37}{3}\text{m}$