Q.Four position-time ($x$ versus $t$) graphs of a particle moving along a straight line are described below. In only ONE of them can the average velocity over the interval $(0, T)$ be made to vanish for a suitably chosen $T$. Which one is it?
Graph (a): at $t=0$ the position is negative (below the $t$-axis); the curve rises, crosses $x=0$ at some later time, reaches a positive maximum and then falls slightly, remaining 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.
Graph (c): at $t=0$ the position is a large positive value; the curve decreases steadily (concave up) and levels off toward zero, staying positive.
Graph (d): the curve starts at the origin ($x=0$ at $t=0$), rises quickly and then flattens (concave down), approaching a constant positive value.

Graph (b) is non-monotonic: it goes up and then comes back down past the level it started from, so there exists a $T>0$ with $x(T)=x(0)$ and hence $\bar v=0$. Graph (a) starts negative and ends positive (never returns to its negative start); graphs (c) and (d) are monotonic, so their position never repeats.

Concept

Average velocity depends only on the net displacement between the endpoints, not on the path in between:

For $\bar v=0$ we need $x(T)=x(0)$ with $T>0$ — the position–time curve must return to the same height it had at $t=0$.

Checking each graph

• (a) starts at a negative $x$ and rises to a positive value where it stays. It never comes back down to its negative starting value, so $x(T)\neq x(0)$ for any $T>0$.
• (b) starts positive, rises to a peak and then falls back down, passing through its initial height again. At that instant $x(T)=x(0)$, so $\bar v=0$. ✓
• (c) decreases monotonically; $x$ never repeats a value, so it can never equal $x(0)$ again.
• (d) increases monotonically toward a constant; again $x$ never returns to its start.

Only a curve that reverses direction can satisfy $x(T)=x(0)$, and graph (b) is the only one that does.

Concept: A Turning Point Is Necessary for Zero Average Velocity — Rolle's Theorem

Method: Two-Part Theorem-Based Test (Rolle's Theorem to eliminate, Intermediate Value Theorem to confirm)

Rather than checking, for each graph individually, "does it come back to its starting height," this method uses one calculus theorem to eliminate three of the four graphs at once — by a necessary condition every candidate must satisfy — and a second theorem to positively confirm the survivor.

Steps

1. State the target condition. Average velocity over $(0,T)$ vanishes iff $x(T) = x(0)$ for some $T>0$, since $\bar v = [x(T)-x(0)]/T$.

2. Rolle's Theorem (the elimination step). If $x$ is smooth and $x(T)=x(0)$ for some $T>0$, Rolle's Theorem guarantees there exists an instant $c\in(0,T)$ with $x'(c)=0$ — i.e. an interior turning point (a local max or min of $x$) must exist strictly between $0$ and $T$. Contrapositive: a graph with no interior turning point at all can never satisfy $x(T)=x(0)$ for any $T>0$.

3. Scan all four graphs for an interior turning point:

   • (a) rises steadily from a negative start to a positive plateau (with only a slight late dip that stays positive) — essentially monotonic, no interior turning point that brings it back down to its starting level.
   • (b) rises to a peak, then falls — has an interior turning point (the peak).
   • (c) decreases steadily, levelling off — monotonic, no turning point.
   • (d) rises and flattens toward a constant — monotonic, no turning point.

   By Step 2's contrapositive, (a), (c), and (d) are eliminated immediately — none of them can ever return to its starting value, so $\bar v=0$ is impossible for any $T$ on those three.

4. Intermediate Value Theorem (the confirmation step, for graph (b) only). Graph (b) starts at $x(0)>0$, rises above $x(0)$ to its peak, then falls in an S-shape to a smaller positive value below $x(0)$. Since $x$ is continuous and takes a value above $x(0)$ (at the peak) and later a value below $x(0)$ (at the end), the Intermediate Value Theorem guarantees $x$ passes through the value $x(0)$ exactly at some intermediate time $T$ during the fall — this is the required $T$ with $x(T)=x(0)$.

Why two theorems, not one

Rolle's Theorem alone only tells you a turning point is necessary — it doesn't by itself prove the curve actually returns to its starting height (a curve could turn around and still never come back down that far). It's the Intermediate Value Theorem, applied specifically to graph (b)'s stated overshoot-then-undershoot behaviour, that positively confirms the crossing exists.

Final Answer

— the only one with an interior turning point (necessary, by Rolle's Theorem) that is also confirmed, by the Intermediate Value Theorem, to bring $x$ back through its starting value.