Refer to the graph in figure. 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 $t > 0$
$(D)$	$(iv)$ has $v < 0$ and $a > 0$

(A-III, B-II, C-IV, D-I) We analyze the slope of each curve,which represents velocity $v = \frac{dx}{dt}$,and the curvature,which relates to acceleration $a = \frac{d^2x}{dt^2}$.
$1$. Graph $(A)$: The curve crosses the $t$-axis at point $B$,meaning displacement $x = 0$ for some $t > 0$. Thus,$(A)$ matches $(iii)$.
$2$. Graph $(B)$: The curve stays above the $t$-axis $(x > 0)$. It has a peak at $B_1$ where slope $v = 0$. It also has a point of inflection between $B_1$ and $C_1$ where the curvature changes,meaning $a = 0$. Thus,$(B)$ matches $(ii)$.
$3$. Graph $(C)$: The slope is negative throughout $(v < 0)$ and the curve is concave up (slope becomes less negative),meaning $a > 0$. Thus,$(C)$ matches $(iv)$.
$4$. Graph $(D)$: The slope is positive $(v > 0)$ and the curve is concave down (slope decreases),meaning $a < 0$. Thus,$(D)$ matches $(i)$.
Final matching: $(A)-(iii), (B)-(ii), (C)-(iv), (D)-(i)$.