Q.For the one-dimensional motion, described by $x = t - \sin t$ (Note: more than one of the given options may be correct.)

Concept: One Dimensional Motion — position, velocity, acceleration from $x(t)$.

Step 1 — Velocity and acceleration

$x = t - \sin t$

$v = \frac{dx}{dt} = 1 - \cos t$

$a = \frac{dv}{dt} = \sin t$

Step 2 — Check each option

(A) For $t>0$, $t - \sin t > 0$ (since $t > \sin t$ for $t>0$). True. …

Let’s understand the motion physically before diving into algebra. The equation $x = t - \sin t$ combines a steady drift ($t$) with a periodic wiggle ($-\sin t$). Imagine a point that moves uniformly to the right but also oscillates back and forth — the net effect is that it never goes backward, but its speed varies.

We need to check each statement carefully. Since more than one option may be correct, we treat each independently.

1. Check option (A): $x(t) > 0$ for all $t > 0$.

   For $t > 0$, we have $x = t - \sin t$. The sine function satisfies $\sin t \leq 1$, so $t - \sin t \geq t - 1$. For $t > 1$, this is clearly positive. For $0 < t \leq 1$, note that $\sin t < t$ for all $t > 0$ (a standard inequality: the sine curve lies below its tangent at the origin). Hence $t - \sin t > 0$ for every $t > 0$. At $t=0$, $x=0$, but the statement says "for all $t > 0$", so it holds.

   Option (A) is correct.

2. Check option (B): $v(t) > 0$ for all $t > 0$.

   Velocity is the derivative: $v = \frac{dx}{dt} = 1 - \cos t$. Since $\cos t$ ranges from $-1$ to $1$, $1 - \cos t$ ranges from $0$ to $2$. It equals $0$ whenever $\cos t = 1$, i.e., at $t = 2\pi n$ for integer $n$. For $t > 0$, the first such instant is $t = 2\pi$, where $v = 0$. So $v(t)$ is not strictly positive for all $t > 0$ — it becomes zero periodically.

   Option (B) is false.

3. Check option (C): $a(t) > 0$ for all $t > 0$.

   Acceleration is $a = \frac{dv}{dt} = \sin t$. The sine function is positive for $0 < t < \pi$, negative for $\pi < t < 2\pi$, and so on. So $a(t)$ changes sign repeatedly. It is not always positive.

   Option (C) is false.

4. Check option (D): $v(t)$ lies between 0 and 2. …

Concept: Recognising $x=t-\sin t$ as the Cycloid — the Path of a Point on a Rolling Wheel

Method: The Rolling-Wheel Physical Model (derive the velocity bounds from "rolling without slipping," not from bounding trigonometric functions algebraically)

This exact function is not an arbitrary formula — $x(\theta)=R(\theta-\sin\theta)$ (with $y(\theta)=R(1-\cos\theta)$) is the classical cycloid: the horizontal coordinate traced by a point fixed to the rim of a wheel of radius $R$ rolling without slipping along the ground, with $\theta$ the wheel's rotation angle. Here $R=1$ and $\theta=t$ (angular speed $1\ \text{rad s}^{-1}$, so the wheel's centre also moves at speed $1\ \text{m s}^{-1}$). Recognising this turns options (b)–(d) into a direct application of the well-known physics of rolling without slipping, rather than a bare calculus exercise.

Steps

1. Differentiate to confirm the standard formulas (needed regardless of interpretation):

2. Recognise the physical model. A wheel of radius $1$ rolling without slipping, centre moving at constant speed $1\ \text{m s}^{-1}$, has its rim point's horizontal velocity equal to $v_{\text{centre}} + v_{\text{rotation, horizontal component}} = 1 + 1\cdot(-\cos t) = 1-\cos t$ — exactly matching Step 1's derivative, confirming the identification.

3. (D) — the velocity bound, from rolling geometry, not algebra. The defining fact of rolling without slipping is that the point of the wheel touching the ground is, at that instant, momentarily at rest relative to the ground (this is literally what "no slipping" means — otherwise the wheel would be skidding). That contact point occurs once per rotation, at $\theta = 2\pi n$, i.e. $\cos t=1 \Rightarrow v=1-1=0$ — the minimum of $v$. The maximum occurs at the top of the wheel, diametrically opposite the contact point ($\theta=\pi$, $\cos t=-1$), where a rim point's speed is the well-known result "twice the centre's speed" $=2\times 1 = 2$. So $v$ ranges over exactly $[0,2]$ — read directly from wheel geometry, without needing to bound $\cos t\in[-1,1]$ algebraically. (D) is correct.

4. (B) — must $v>0$ for all $t>0$? Since the contact-point velocity is exactly zero once every rotation ($t=2\pi, 4\pi,\ldots$), $v$ touches (but never goes below) zero periodically — it is never negative (a rolling-without-slipping point never moves backward relative to the ground) but it is not strictly positive at every instant. (B) is false. …