Refer to the graphs fig. 2 (EP).5 Match the following.
Graph Characteristic
(a) (i) has ` vgt and a lt o` throughout.
(b) (ii) has ` xgt 0` throughout and has a point with ` v=0` and a point with `a=0`.
(c ) (ii) has a point with zero displacement for ` tgt 0`.
(d) (iv) has ` v lt and a gt 0`.

Explanation

`(a-iii), (b-ii), (c-iv), (d-i)`.
For graph (a). There is a point (B) for which displacementis zero for ` t gt0`, which is option (iii). For graph (b), the condition ` x gt 0` throughout. There is a point ` B_(1)` at the top of the curve, where slope of the curve, i.e., ` dx//dt=v=0` and there is a point ` C_(1)` on curve, wher the curvature of curve chages, which show, acceleration=0. This corresponds to option (ii).
For graph (c ), the slope (dx//dt) is negative, hence ` v lt0`. The curvature of ` x-t` graph indicates that the begative velocity (=dx//dt) is decreasing with time. Therefore, ` a gt0`. This corresponds to option (iv).
For graph (d), the slope (=dx//dt) is poitive, hence ` vgt0`. The curvature of ` x-t` graph indicates that positive velocity is decreasing with time. Therfore, ` a lt 0`. This corresponds to option (i).