Question

Graph of $\frac{1}{v}vs.x$ for a particle under motion is as shown, where $v$ is velocity and $x$ is position. The time taken by particle to move from $x=4mtox=12m$ is,

• A. 12 $sec$
• B. 10 $sec$
• C. 8 $sec$
• D. $\frac{16}{3}$ $sec$

Answer 1

The correct option is B 10 $sec$

${\int }_{t_1}^{t_2}dt=$Area under graph
between $4m$ and $12m$
As we know, $v=\frac{dx}{dt}$
$\Rightarrow \int dt=\int 1\frac{dx}{v}=\int 1v^{-1}dx$
$\Rightarrow {\int }_{t_1}^{t_2}dt=$Area under graph between
4 $m$ and $12m$

Area of graph $=$ Area of trapezium
$=\frac{1}{2}(\frac{1}{2}+2)\times (12-4)$
$\therefore t_2-t_1=\frac{1}{2}\times \frac{5}{2}\times 8=10s$
Time taken =10 $s$.
Final Answer: $b$