Question

Graph of (1/v) vs x for a particle under motion is as shown, where v is velocity x is position. The time taken by particle to move form x=4m to x=12 m is

• A. $16sec$
• B. $10sec$
• C. $8sec$
• D. $12sec$

Answer 1

The correct option is B $10sec$

We know

$v=\frac{dx}{dt}$

$\Rightarrow \int dt=\int \frac{dx}{v}=\int v^{-1}dx$

$\Rightarrow {\int }_{t_1}^{t_2}dt=$Area under graph between $z=4m$ & $12m$

Area of graph$=$Area of trapezium

$=\frac{1}{2}(\frac{1}{2}+2)\times (12-4)$

$t_2-t_1=\frac{1}{2}\times \frac{5}{2}\times 8=10s$

Time taken$=10s$.