Q.In one dimensional motion, instantaneous speed $v$ satisfies $0 \le v < v_0$.

Concept: Speed Bounds Only the Magnitude of Velocity

Speed $v = |v_x|$, so $0 \le v < v_0$ means $-v_0 < v_x < v_0$ — the velocity can be positive, negative, or zero at any instant; only its magnitude is restricted.

Step 1 — (B) holds. $x = \int_0^T v_x\,dt$, and $|x| \le \int_0^T |v_x|\,dt < v_0 T$, so $-v_0T < x < v_0T$ always.

Step 2 — (A) fails. If the particle moves backward the whole time (e.g. $v_x = -v_0/2$ constant), $x(T) = -v_0T/2 < 0$.

Step 3 — (C) fails. A particle moving forward while decelerating (e.g. $v_x$ decreasing from $v_0/2$ toward $0$) has $a = dv_x/dt < 0$.

Step 4 — (D) fails. Since $v_x = 0$ is allowed and direction is unconstrained, the particle can stop and reverse — a turning point (e.g. $v_x = \frac{v_0}{2}\cos(\pi t/T)$).

Setting Up: Speed Is Not Velocity

In one dimension, let $v_x$ be the (signed) instantaneous velocity of the particle, so its speed is $v = |v_x|$. The given condition $0 \le v < v_0$ means:

At every instant $v_x$ can be positive, negative, or zero — the bound restricts only how large $|v_x|$ can get, not which direction the particle moves.

Statement (B): the only one that must hold

Displacement over time $T$ is $x = \int_0^T v_x \, dt$. Using the bound $|v_x| < v_0$ pointwise:

So $-v_0 T < x < v_0 T$ for every possible motion satisfying the given speed bound. (B) is always true.

Statement (A): displacement always non-negative — FALSE

Counter-example: let the particle move only in the negative direction throughout, $v_x(t) = -\dfrac{v_0}{2}$ (constant, satisfies $|v_x| < v_0$). Then

which is negative, so (A) does not always hold.

Statement (C): acceleration always non-negative — FALSE

Counter-example: let the particle move in the positive direction while slowing down, e.g. $v_x(t) = \dfrac{v_0}{2}\left(1 - \dfrac{t}{2T}\right)$ for $0 \le t \le T$ (this stays inside $-v_0 < v_x < v_0$). Then

a negative acceleration, so (C) does not always hold.

Statement (D): no turning points — FALSE

The bound $0 \le v < v_0$ explicitly permits $v = 0$, i.e. $v_x = 0$ is allowed, and there is no restriction preventing the particle from moving forward, coming to rest, and then moving backward. For example, $v_x(t) = \dfrac{v_0}{2}\cos\left(\dfrac{\pi t}{T}\right)$ satisfies $|v_x| \le v_0/2 < v_0$ and changes sign at $t = T/2$ — a genuine turning point. So (D) does not always hold.

Conclusion

Only the bound on displacement, statement (B), is forced by the given condition on speed; (A), (C), and (D) can each be violated.

Concept: Visualizing the Constraint as a Confined Strip in the $v$-$t$ Plane

Method: The Horizontal-Strip Picture (one geometric device tests all four statements at once)

Rather than constructing a separate explicit counter-example function for each of the four statements, this method turns the given condition into a single picture — a horizontal strip in the $v_x$-$t$ plane — and reads all four answers directly off what that picture does and does not allow.

Setting up the picture

1. Translate the speed bound into a velocity bound. Speed $v = |v_x|$, so $0\le v<v_0$ means

Plotted against $t$, this describes an open horizontal strip of total width $2v_0$, centred on $v_x=0$: the particle's velocity-time curve $v_x(t)$ must stay strictly inside this strip at all times, but is otherwise completely unconstrained — it can wander anywhere within the strip, with any shape, any slope, any number of zero-crossings.

Reading each statement off the strip

2. (B) — the displacement bound. Displacement is the signed area under the $v_x(t)$ curve, $x=\int_0^T v_x\,dt$. Since the curve never leaves the strip $(-v_0,v_0)$, the area under it over a duration $T$ is bounded, in magnitude, by the area of the strip itself over that same duration: $|x| < v_0 T$, i.e. $-v_0T < x < v_0T$. This holds for any curve that stays inside the strip, regardless of its shape — so (B) is always true.

3. (A) — displacement always non-negative? The strip extends symmetrically below zero as well as above; a curve can lie entirely in the lower half of the strip the whole time (e.g. a flat line at $v_x=-v_0/2$, well inside the strip). Such a curve gives $x(T)<0$. (A) is false — the strip picture permits it directly.

4. (C) — acceleration always non-negative? The strip constrains only the vertical position of the curve (how large $|v_x|$ can get) — it says nothing whatsoever about the curve's slope. A curve can slope downward while still staying inside the strip (e.g. drifting from $v_0/2$ down toward $0$, still positive, but decreasing) — that's a particle moving forward while decelerating, i.e. $a<0$. (C) is false.

5. (D) — no turning points? The value $v_x=0$ lies well inside the open strip (it's the strip's own centre line), so a curve is free to cross it — any number of times, e.g. an oscillating curve confined to the strip. Crossing $v_x=0$ means reversing direction — a turning point. (D) is false.

Why one picture suffices for all four

Every statement reduces to a question about what a curve confined to this specific strip is or isn't forced to do — the area under it (bounded, giving (B) true), where it sits vertically (either half, giving (A) false), its slope (totally free, giving (C) false), and whether it must avoid the centre line (it needn't, giving (D) false). No separate counter-example construction is needed statement by statement; the strip picture answers all four questions from one setup.

Final Answer