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Refer to the graphs given below. Match the following.

Graph Characteristics
(a) (i) has v > 0 and a < 0 throughout.
(b) (ii) has x > 0 throughout and has a point with v = 0 and a point with a = 0.
(c) (iii) has a point with zero displacement for > 0.
(d) (iv) has v < 0 and a > 0

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(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.

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