Question

The $\frac{1}{V}$ (V : velocity) versus displacement graph of a particle is shown in the figure. The particle is moving in a straight line along positive x-axis, then

Answer 1

$v=\frac{x}{t}$
$t=\frac{x}{v}=x\times \left(\frac{1}{v}\right)$
$\begin{array}{cc}& dt=\left(dx\right)\frac{1}{v}\\ & t=\int \frac{1}{v}dx\end{array}$
Area of $\frac{1}{v}$ versus $x$ graph

$\begin{array}{cc}& t=\frac{1}{2}(2+4)\times 2=6sec\\ & \text{Equation of line,}\\ & \frac{1}{v}=2+x\\ & v=\frac{1}{(2+x)}\\ & \therefore a=v\frac{dv}{dx}\\ & =\left(\frac{1}{2+x}\right)[-{\left(\frac{1}{2+x}\right)}^2]\\ & a=-{\left(\frac{1}{2+x}\right)}^3\\ & \text{acceleration of point A}\\ & x=0\\ & a=-{\left(\frac{1}{2}\right)}^3\\ & a=-\frac{1}{8}m/{\text{sec}}^2\end{array}$