Particle moving in a straight line is shown in fig. Corresponding acceleration versus velocity graph will be 

detailed solution

Correct option is A

acceleration  a=dvdt                    acceleration→a=dvdsdsdt                                             a=dvds.V..........(1)     ∵dsdt=VFrom given graph slope of given line in graph=dvds=difference of velocity coordinatesdifference of displacement coordinates dvds=10-010-0=1010=1 S−1---(2)∴ From Equation (1)   and (2)   acceleration→a=1×V  (from above)  ⇒a=VTherefore acceleration versus velocity graph will be a straight line passing through origin with slope  1 S−1.