(a) – (iii) From graph, at point B, x = 0 For t > 0. Hence (a) → (iii)
(b) – (ii) From graph, throughout the motion x > 0. At point C, \(\frac{dx}{dt}\) = v = 0
Since at D, curvature changes, hence a = 0
(c) – (iv) From graph, \(\frac{dx}{dt}\) = negative ; So, v < 0; a = \(\frac{d^2x}{dt^2}\) is positive.
Since rate of change of negative velocity decreases. So, a > 0
(d) – (i) This graph is reverse of the graph above v > 0 but a < 0.