\(v = - mx + {v_0}\)…..(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 \(\frac{{dv}}{{dt}} = - m[ - mx + {v_0}] = {m^2}x - m{v_0}\)

\(\therefore a = {m^2}x - m{v_0}\)

i.e. the graph between \(a\) and \(x\) should have positive slope but negative intercept on \(a-\)axis. So graph \((a) \) is correct.