Question

The velocity displacement graph of a particle moving along a straight line is shown

The most suitable acceleration -displacement $(a-x)$ graph will be

• A.
• B.
• C.
• D.

Answer 1

The correct option is A

$v=-\frac{v_0}{x_0}x+v_0$,
Since we know $a=v\frac{dv}{dx}$ $\Rightarrow a=(-\frac{v_0}{x_0}x+v_0)\left(\frac{-v_0}{x_0}\right)$
$\Rightarrow a=(\frac{v_0}{x_0}{)}^{2}x-\frac{v_0^2}{x_0}$
This is the equation of straight with positive slope and negative intercept.
$\therefore$ So the correct option is $\left(A\right)$ and the graph is