In This post you will get answer of Q.
The velocity-displacement graph of a particle is as shown in the figure. Which of the following graphs correctly represents the variation of acceleration with displacement ?
Solution:
For the given velocity-displacement graph, intercept $=v_{0}$ and slope $=-frac{v_{0}}{x_{0}}$
Thus, the equation of given line of velocity-displacementgraph is
$v=frac{v_{0}}{x_{0}}x+v_{0}quadldotsleft(iright)$
Acceleration, $a=frac{dv}{dt}=frac{dv}{dx} frac{dx}{dt}=frac{dv}{dx}v$
$because frac{dv}{dx}=-frac{v_{0}}{x_{0}}$
$therefore a=frac{v_{0}}{x_{0}}left(-frac{v_{0}}{x_{0}}x+v_{0}right)quad$ (Using $(i)$)
$=frac{v^{2}_{0}}{x^{2}_{0}}x-frac{v^{2}_{0}}{x_{0}}$
It is a straight line with positive slope and a negative intercept.
The variation of $a$ with $x$ is as shown in the above figure.
Solution:
For the given velocity-displacement graph, intercept $=v_{0}$ and slope $=-frac{v_{0}}{x_{0}}$
Thus, the equation of given line of velocity-displacementgraph is
$v=frac{v_{0}}{x_{0}}x+v_{0}quadldotsleft(iright)$
Acceleration, $a=frac{dv}{dt}=frac{dv}{dx} frac{dx}{dt}=frac{dv}{dx}v$
$because frac{dv}{dx}=-frac{v_{0}}{x_{0}}$
$therefore a=frac{v_{0}}{x_{0}}left(-frac{v_{0}}{x_{0}}x+v_{0}right)quad$ (Using $(i)$)
$=frac{v^{2}_{0}}{x^{2}_{0}}x-frac{v^{2}_{0}}{x_{0}}$
It is a straight line with positive slope and a negative intercept.
The variation of $a$ with $x$ is as shown in the above figure.