Q.Four position-time ($x$ versus $t$) graphs of a particle moving along a straight line are described below.
Graph (a): at $t=0$ the position is negative (below the $t$-axis); the curve rises, crosses $x=0$ at some later time $t>0$, reaches a positive maximum and then falls slightly, staying positive.
Graph (b): at $t=0$ the position is positive; the curve rises to a maximum and then falls in an S-shape to a smaller positive value, remaining positive throughout.
Graph (c): at $t=0$ the position is a large positive value; the curve decreases steadily and is concave up (its downward slope becomes gentler), levelling off toward zero while staying positive.
Graph (d): the curve starts at the origin, rises quickly and then flattens (concave down), approaching a constant positive value.
Match each graph to its characteristic:

Analysing each graph

• Graph (a): starts negative and rises through the $t$-axis, so there is a time $t>0$ at which $x=0$ — a point of zero displacement for $t>0$ $\Rightarrow$ (iii).
• Graph (b): stays positive throughout; it rises to a maximum (slope $=0$, so $v=0$ there) and then descends through an inflection (curvature $=0$, so $a=0$ there). It has $x>0$ throughout with both a $v=0$ point and an $a=0$ point $\Rightarrow$ (ii). …

Concept: Reading $v$ and $a$ Off an $x$-$t$ Graph via Finite Differences on a Concrete Numeric Model

Method: Build a Numeric Table for Each Graph, Verify with Central Differences (not abstract slope/curvature language)

The stored answer reasons about "slope $=v$" and "curvature $=a$" in the abstract. This method instead builds one concrete, self-consistent set of $(t,x)$ values matching each graph's described shape, and computes discrete central differences $v\approx \dfrac{x(t+\Delta t)-x(t-\Delta t)}{2\Delta t}$ and $a\approx x(t+\Delta t) - 2x(t) + x(t-\Delta t)$ — the same numeric cross-check technique used elsewhere in this chapter for sketch-reading, applied here to sign/turning-point questions instead of ranking questions.

Graph (a) — negative start, crosses zero, small positive peak, slight fall

The table shows $x=0$ exactly at $t=1>0$ — a point of zero displacement for $t>0$, directly from the values, no differencing needed. $\Rightarrow$ (iii).

Graph (b) — positive start, rises to a max, falls in an S-shape, stays positive

All values stay positive ($x>0$ throughout). Central difference at the symmetric peak: $v(2) \approx \dfrac{x(2.5)-x(1.5)}{2}$; choosing $x(1.5)=x(2.5)=4.5$ symmetric about the peak $x(2)=5$ gives $v(2)\approx\dfrac{4.5-4.5}{2}=0$ — a genuine turning point, $v=0$. For the inflection, compare second differences on the falling side: $a(3)\approx x(4)-2x(3)+x(2) = 3.8-9+5=-0.2$ (still concave down) but $a(4)\approx x(5)-2x(4)+x(3)=3.3-7.6+4.5=+0.2$ (now concave up) — a sign change in the second difference between $t=3$ and $t=4$ confirms a point with $a=0$ somewhere in between. Both a $v=0$ point and an $a=0$ point, with $x>0$ throughout. $\Rightarrow$ (ii).

Graph (c) — large positive start, decreasing, concave up, levelling toward zero

Central differences: $v(1)\approx\dfrac{7-20}{2}=-6.5$, $v(2)\approx\dfrac{4-12}{2}=-4$, $v(3)\approx\dfrac{2.5-7}{2}=-2.25$ — all negative, matching $v<0$. Second differences: $a(1)\approx7-24+20=3$, $a(2)\approx4-14+12=2$, $a(3)\approx2.5-8+7=1.5$ — all positive, matching $a>0$ (the successive drops $-8,-5,-3,-1.5$ shrink in magnitude, the numeric signature of concave-up). $\Rightarrow$ (iv).

Graph (d) — starts at origin, rises quickly, flattens, concave down