Explanation

(a) Given line have positive intercept but negative slope. So its equation can be written as
$v = - mx + v_{0}$ ….(i) [where $m = \tan\theta = \frac{v_{0}}{x_{0}}$]
By differentiating with respect to time we get
$\frac{dv}{dt} = - m\frac{dx}{dt} = - mv$
Now substituting the value of $v$ from eq. (i) we get
$\frac{dv}{dt} = - m\left\lbrack - mx + v_{0} \right\rbrack = m^{2}x - mv_{0}\ \therefore a = m^{2}x - mv_{0}$
$i.e.$ the graph between $a$ and $x$ should have positive slope but negative intercept on $a$-axis. So graph (a) is correct