The slope of the given v versus x graph is $$m = - {{{v_0}} \over {{x_0}}}$$ and intercept is c = + v₀. Hence, v varies with x as

$$v = - \left( {{{{v_0}} \over {{x_0}}}} \right)x + {v_0}$$ ...... (1)

where v₀ and x₀ are constants of motion. Differentiating with respect to time t, we have

$${{dv} \over {dt}} = - \left( {{{{v_0}} \over {{x_0}}}} \right){{dx} \over {dt}}$$

or $$a = - \left( {{{{v_0}} \over {{x_0}}}} \right)v$$ ........ (2)

Using Eq. (1) in Eq. (2), we get

$$a = - \left( {{{{v_0}} \over {{x_0}}}} \right)\left( { - {{{v_0}} \over {{x_0}}}x + {v_0}} \right)$$

$$a = {\left( {{{{v_0}} \over {{x_0}}}} \right)^2}x - {{v_0^2} \over {{x_0}}}$$

Thus the graph of a versus x is a straight line having a positive slope = $${\left( {{{{v_0}} \over {{x_0}}}} \right)^2}$$ and negative intercept = $$ - {{v_0^2} \over {{x_0}}}$$. Hence the correct choice is (a).