Q.Read each statement below carefully and state with reasons and examples, if it is true or false; A particle in one-dimensional motion

Concept: Instantaneous Velocity — speed is the magnitude of velocity; acceleration is the rate of change of velocity, not of speed.

1. True. Speed is $|\vec{v}|$. At an instant, a particle can have zero speed (e.g., at the turning point of a vertical throw) while its velocity is changing direction — so acceleration $a = \frac{dv}{dt} \neq 0$. Example: A ball thrown upward has $v=0$ at the highest point, but $a = -g \neq 0$.
2. False. Speed is the magnitude of velocity. If speed is zero, then $|\vec{v}| = 0$, which forces $\vec{v} = \mathbf{0}$. Zero speed implies zero velocity.
3. False. Constant speed means $|\vec{v}|$ is constant, but the direction of velocity can change — that requires acceleration. …

The Core Idea: Speed, Velocity, and Acceleration

Before we judge each statement, we need a crystal-clear picture of three quantities in one-dimensional motion.

Velocity is a vector — it has both magnitude and direction. In one dimension, we represent direction with a sign: $+$ for one way, $-$ for the opposite. Speed is the magnitude of velocity, always non-negative. Acceleration is the rate of change of velocity, not of speed.

This last point is the key that unlocks all four statements. A particle can have a large velocity that is decreasing (negative acceleration), or zero velocity but a non-zero rate of change (acceleration). Let's examine each case.

Step-by-Step Analysis

1. Statement (a): "With zero speed at an instant may have non-zero acceleration at that instant"

Think of a ball thrown straight upward. At the very top of its path, it stops for an instant — its speed is zero. But gravity is still pulling it downward, so its velocity is changing from positive (upward) to negative (downward). That change means acceleration is non-zero ($-g$).

Mathematically: if $v(t) = 0$ at some instant $t_0$, then $a(t_0) = \frac{dv}{dt}\big|_{t_0}$ can be anything — it depends on the slope of the velocity-time graph at that point, not on the value of $v$ itself.

So statement (a) is True.

2. Statement (b): "With zero speed may have non-zero velocity"

Speed is defined as $|v|$. If speed is zero, then $|v| = 0$, which forces $v = 0$. There is no way for a vector to have magnitude zero but a non-zero value — that would be like saying a distance of zero metres is actually 5 metres.

So statement (b) is False.

3. Statement (c): "With constant speed must have zero acceleration"

Constant speed means $|v|$ is constant. But velocity can still change direction. In one-dimensional motion, "direction" is just sign — so constant speed with changing velocity means the particle is moving back and forth, like a ball bouncing between two walls with the same speed.

Consider a particle moving with velocity $v = +5 \, \text{m/s}$ for one second, then instantly reversing to $v = -5 \, \text{m/s}$ for the next second. Its speed is always $5 \, \text{m/s}$, but at the instant of reversal, acceleration is theoretically infinite (in practice, very large). Even without an instant reversal, consider uniform circular motion projected onto one dimension — that's simple harmonic motion, where speed is constant only at specific points, but acceleration is never zero.

Concept: Testing Kinematic Claims with Explicit Constructed Functions

Method: Proof by Explicit Function Construction

Rather than reasoning narratively about each statement, this method constructs one explicit, simple $v(t)$ (or notes when no such function can possibly exist) for every part, and checks the claim by direct substitution — a technique that generalises to any true/false kinematics claim.

Steps

1. (a) "Zero speed may have non-zero acceleration." Construct $v(t) = 20 - 10t\ \text{m s}^{-1}$ (a ball thrown up at $20\ \text{m s}^{-1}$, $g=10$). At $t=2\ \text{s}$: $v(2)=0$ (zero speed). But $a = dv/dt = -10\ \text{m s}^{-2} \ne 0$ at every instant, including $t=2$. The explicit function exhibits zero speed and non-zero acceleration simultaneously.

   → True.

2. (b) "Zero speed may have non-zero velocity." This one admits no counterexample, by the definition of speed: $\text{speed} = |v|$. If speed $=0$ then $|v|=0$, and the only real number with magnitude zero is $v=0$ itself — there is no function $v(t)$ with $|v(t_0)|=0$ and $v(t_0)\ne 0$ simultaneously, because that would require a nonzero number whose magnitude is zero. Unlike the other three parts, this is settled by definition, not by example.

   → False.

3. (c) "Constant speed must have zero acceleration." Construct a piecewise $v(t)$: $v=+5\ \text{m s}^{-1}$ for $0<t<T$, then $v=-5\ \text{m s}^{-1}$ for $T<t<2T$ (a particle bouncing between two walls). Speed $|v(t)|=5\ \text{m s}^{-1}$ is constant for every $t$ in the domain, yet at $t=T$ the velocity jumps from $+5$ to $-5$ — a sharp, non-zero (in fact very large) acceleration at the bounce, even though speed never changed.

   → False.

4. (d) "Positive acceleration must mean speeding up." Construct $v(t) = -10 + 2t\ \text{m s}^{-1}$ (constant $a=+2\ \text{m s}^{-2}$, particle initially moving in the $-x$ direction). Tabulate:

   | $t$ (s) | $0$ | $1$ | $2$ | $5$ | …