Question

The given graph shows the variation of velocity with displacement. Which one of the graph given below correctly represents the variation of acceleration with displacement ?

• A.
• B.
• C.
• D.

Answer 1

The correct option is B

From the $v-x$ graph, we can do an analysis for acceleration using
$a=v\frac{dv}{dx}=v$ (slope of tangent in $v-x$ graph)

From the graph, we have $v\left(x\right)=\left(\frac{-v_0}{x_0}\right)x+v_0$

Slope of tangent $=\frac{dv}{dx}=\frac{-v_0}{x_0}$

This slope is constant as $v-x$ graph is a straight line.
$\Rightarrow a=v(-\frac{v_0}{x_0})$

$\Rightarrow a\left(x\right)=(\frac{-v_{0}x}{x_0}+v_0)\left(\frac{-v_0}{x_0}\right)=\frac{v_0^2}{x_0^2}x-\frac{v_0^2}{x_0}$

$\Rightarrow a-x$ graph is a striaght line with slope $\frac{v_0^2}{x_0^2}$ and intercept on $Y-$ axis equal to $\frac{-v_0^2}{x_0}$

$\Rightarrow$ (a) is correct answer.

Alternative Approach:
$\left(i\right)$ At $x=0,a=v\frac{dv}{dx}=v_0\left(\frac{-v_0}{x_0}\right)=-ve$

$\left(ii\right)$ At $x=x_0,a=v\frac{dv}{dx}=0\left(\frac{-v_0}{x_0}\right)=0$

Hence, option (a) is the only possible answer.