For the one-dimensional motion, described by $x = t - \sin t$ (Note: more than one of the given options may be correct.) (a) $x(t) > 0$ for all $t > 0$. (b) $v(t) > 0$ for all $t > 0$. (c) $a(t) > 0$ for all $t > 0$. (d) $v(t)$ lies between 0 and 2.

> [!TLDR] > The position function $x = t - \sin t$ describes a particle that always moves forward with positive displacement for $t>0$, but its velocity oscillates between 0 and 2, and acceleration alternates sign. Only options (A) and (D) are correct. 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 $0 0$ (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.** …

For the one-dimensional motion, described by x = t – Class 11 Physics

NCERT Exemplar · Q9

x = t - \sin t

(a)

x(t) > 0

for all

t > 0

(b)

v(t) > 0

for all

t > 0

(c)

a(t) > 0

for all

t > 0

(d)

v(t)

lies between 0 and 2.

Andhra Pradesh BieapMCQ· 1mImportance★★★★★est

67% · 34/51 Questions

🔒 Locked · start free trial →

You're viewing a preview — the full solution, concept, methods & PYQ mapping are locked.

Start your 14-day free trial to unlock the full solution →

Concept understanding — One Dimensional Motion

One Dimensional Motion

Imagine you're standing on a long, perfectly straight railway track. A train moves along it — it can only go forward or backward. It cannot turn left, right, up, or down. That's the core idea: motion confined to a single straight line.

The Intuition

In the real world, a ball thrown across a room moves in three dimensions — it goes forward, sideways, and up-down. But many problems in physics are simpler. We deliberately restrict motion to one dimension (1D) to understand the fundamental laws without the clutter of angles and curves.

Think of:

A car moving on a straight highway (no turns).
A lift going up or down a shaft.
A ball dropped straight down from a height.
A puck sliding on a frictionless straight track.

In each case, the object's position can be described by just one number — its distance from a fixed point (the origin) along that line.

The Precise Statement

x

This means:

The path is a straight line.
The direction is either positive (say, to the right or upward) or negative (left or downward).
All vector quantities (displacement, velocity, acceleration) have only two possible directions — forward or backward.

The Three Key Quantities

To describe 1D motion precisely, we use three quantities:

Position ( $x$ or $y$ ) — where the object is relative to the origin.

Example: $x = +5\ \text{m}$ means 5 metres to the right of the origin.
Displacement ( $\Delta x$ ) — change in position:

\Delta x = x_{\text{final}} - x_{\text{initial}}

This is a vector — it has a sign. If you move from $x=2\ \text{m}$ to $x=7\ \text{m}$ , $\Delta x = +5\ \text{m}$ . If you move back to $x=3\ \text{m}$ , $\Delta x = -4\ \text{m}$ .

Velocity ( $v$ ) — rate of change of position:

v = \frac{\Delta x}{\Delta t}

Average velocity has a sign. Instantaneous velocity is the slope of the position-time graph.

Acceleration ( $a$ ) — rate of change of velocity:

a = \frac{\Delta v}{\Delta t}

Again, a signed quantity. Positive acceleration doesn't always mean speeding up — it means velocity is becoming more positive (or less negative).

The Equations of Motion (Constant Acceleration)

For the special (and very common) case of constant acceleration, we have three equations that connect these quantities. They are the equations of motion for 1D:

v = u + at

Where:

$u$ = initial velocity
$v$ = final velocity
$a$ = constant acceleration
$t$ = time
$s$ = displacement

Watch out

These equations only work when acceleration is constant. If acceleration changes, you cannot use them directly — you'd need calculus or graphical methods.

A Simple Example …

Start your 14-day free trial to unlock the full solution →

The position function $x = t - \sin t$ describes a particle that always moves forward with positive displacement for $t>0$ , but its velocity oscillates between 0 and 2, and acceleration alternates sign. Only options (A) and (D) are correct.

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.

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

Start your 14-day free trial to unlock the full solution →

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

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

v(t) = \frac{dx}{dt} = 1-\cos t, \qquad a(t) = \frac{dv}{dt} = \sin t

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.
(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.
(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. …

Start your 14-day free trial to unlock the full solution →

Same / Similar Concept — real previous-year questions on the same or a closely similar concept, not this exact question.

AP EAPCET 2026Set eng-2026-05-12-FN1 markMCQ
$t = Px^2 + Qx$
›Reveal solutionSolution
With $t$ given as a function of $x$ , velocity is $1/(dt/dx)$ ; differentiating again and applying the chain rule gives acceleration $=\dfrac{-2P}{(2Px+Q)^3}$ .

Concept and Intuition

When time is expressed as a function of position rather than the usual $x(t)$ , we invert the roles: velocity is the reciprocal of $dt/dx$ , and acceleration must be found via the chain rule $a=\dfrac{dv}{dt}=\dfrac{dv}{dx}\cdot\dfrac{dx}{dt}=\dfrac{dv}{dx}\cdot v$ .

Step-by-Step Solution

$t=Px^2+Qx\Rightarrow \dfrac{dt}{dx}=2Px+Q$ .

Velocity: $v=\dfrac{dx}{dt}=\dfrac{1}{2Px+Q}=(2Px+Q)^{-1}$ .

Differentiate $v$ with respect to $x$ : $\dfrac{dv}{dx}=-1\cdot(2Px+Q)^{-2}\cdot 2P=\dfrac{-2P}{(2Px+Q)^2}$ . …
Start your 14-day free trial to see the full solution →
AP EAPCET 2026Set eng-2026-05-14-AN1 markMCQ
$a = -kv$
›Reveal solutionSolution
Velocity-dependent deceleration leads to exponential decay of velocity; the time to halve the velocity is $\frac{\ln 2}{k}$ — a 'half-life' type result.

Concept and Intuition

When acceleration is proportional to velocity (and opposing it), the velocity decays exponentially with time, exactly like radioactive decay or RC discharge. There's no fixed 'half life' formula to memorize here for arbitrary factors — we derive it directly from the differential equation.

Step-by-Step Solution

Start with $a = \dfrac{dv}{dt} = -kv$ .

Separate variables: $\dfrac{dv}{v} = -k\,dt$ .

Integrate both sides from initial velocity $v_0 = v$ (at $t=0$ ) to $v/2$ (at time $t$ ): $\displaystyle\int_{v}^{v/2} \frac{dv'}{v'} = -k\int_0^t dt'$

LHS: $\ln(v/2) - \ln(v) = \ln(1/2) = -\ln 2$ .

RHS: $-kt$ . …
Start your 14-day free trial to see the full solution →
AP EAPCET 2026Set eng-2026-05-15-AN1 markMCQ
$a = 6t$
›Reveal solutionSolution
This tests integrating a time-varying acceleration to find velocity. The answer is (B), $12\,ms^{-1}$ .

Concept and Intuition

When acceleration is not constant but a known function of time, velocity is obtained by integrating $a(t)$ with respect to time, using the initial condition to fix the constant of integration.

Step-by-Step Solution

Given $a=\dfrac{dv}{dt}=6t$ .

Integrate both sides: $v(t)=\int 6t\,dt = 3t^2+C$ .

Since the particle starts from rest, $v(0)=0 \Rightarrow C=0$ .

So $v(t)=3t^2$ .

At $t=2\,s$ : $v=3(2)^2=3\times4=12\,ms^{-1}$ .

Common Mistakes …
Start your 14-day free trial to see the full solution →
AP EAPCET 2026Set eng-2026-05-15-FN1 markMCQ
$t_1$
›Reveal solutionSolution
Velocity increases on an interval exactly when acceleration is positive there — the trend of the acceleration (rising, falling, or flat) is irrelevant to whether $v$ is increasing, only its sign matters. All three graphs stay positive across $[t_1, t_2]$ , so all three show increasing velocity.

Concept and Intuition

A very common mix-up is to think "acceleration is decreasing, so the object must be slowing down." That confuses the rate of change of velocity with velocity itself. As long as $a(t) = \dfrac{dv}{dt} > 0$ , $v$ is climbing — even if $a$ itself is falling, the velocity is still gaining, just gaining more slowly near $t_2$ than near $t_1$ . The object only starts to slow down once $a$ actually crosses to negative values.

Step-by-Step Solution

Velocity at $t_2$ compared to $t_1$ is given by $v(t_2) = v(t_1) + \displaystyle\int_{t_1}^{t_2} a\,dt$ . The integral (area under the $a$ – $t$ curve) is positive as long as $a(t) \ge 0$ throughout $[t_1, t_2]$ , and it is positive definite (strictly increasing $v$ ) as long as $a(t) > 0$ almost everywhere in that interval.

Graph I: straight line, positive slope, $a>0$ at $t=0$ and growing — clearly $a>0$ on all of $[t_1,t_2]$ . So $v$ increases, and increases at a growing rate. …
Start your 14-day free trial to see the full solution →
AP EAPCET 2024Set ap-2024-05-17-AN1 markMCQ
$t = \sqrt{x} + 3$
›Reveal solutionSolution
Inverting the given $t$ – $x$ relation gives $x=(t-3)^2$ ; differentiating and setting velocity to zero shows the particle is exactly at $x=0$ at that instant.

Concept and Intuition

When a relation is given as $t$ in terms of $x$ rather than the usual $x(t)$ , first invert it to get $x$ as an explicit function of $t$ , then differentiate normally to get velocity as a function of time.

Step-by-Step Solution

$t=\sqrt{x}+3 \Rightarrow \sqrt{x}=t-3 \Rightarrow x=(t-3)^2$ .

$v=\dfrac{dx}{dt}=2(t-3)$ .

$v=0 \Rightarrow t=3$ s.

Substitute back: $x=(3-3)^2=0$ m.

Common Mistakes …
Start your 14-day free trial to see the full solution →
AP EAPCET 2024Set eng-2024-05-21-AN1 markMCQ
$v(x) = 3x^2 - 4x$
›Reveal solutionSolution
Using $a=v\dfrac{dv}{dx}$ with $v=3x^2-4x$ , the acceleration is $(3x^2-4x)(6x-4)$ .

Concept and Intuition

When velocity is given as a function of position rather than time, acceleration is found via the chain rule $a=\dfrac{dv}{dt}=\dfrac{dv}{dx}\cdot\dfrac{dx}{dt} = v\dfrac{dv}{dx}$ — never simply $dv/dx$ alone.

Step-by-Step Solution

$v(x)=3x^2-4x$ .

$\dfrac{dv}{dx} = 6x-4$ .

$a = v\dfrac{dv}{dx} = (3x^2-4x)(6x-4)$ .

Common Mistakes …
Start your 14-day free trial to see the full solution →
AP EAPCET 2024Set eng-2024-05-21-AN1 markMCQ
$a$
›Reveal solutionSolution
The area under the given acceleration-vs-position graph from $x=0$ to $x=1.4$ is $0.40\ \mathrm{m^2/s^2}$ ; combined with $v(0)=0.8$ , this gives $v(1.4)=1.2\ \mathrm{m/s}$ .

Concept and Intuition

Since $a=v\dfrac{dv}{dx}$ , we have $v\,dv = a\,dx$ , so integrating both sides: $\dfrac{v_f^2-v_i^2}{2} = \int a\,dx = \text{area under the } a\text{–}x \text{ graph}$ . This lets us read the answer directly off the graph's geometry without needing $a(t)$ .

Step-by-Step Solution

Break the a–x graph into three regions and find each area:

$x=0$ to $0.4$ : rectangle, $a=0.4$ , area $=0.4\times0.4=0.16$ .

$x=0.4$ to $0.8$ : trapezoid from $a=0.4$ down to $a=0.2$ , width $0.4$ : area $=\dfrac{0.4+0.2}{2}\times0.4=0.12$ .

$x=0.8$ to $1.4$ : rectangle, $a=0.2$ , width $0.6$ : area $=0.2\times0.6=0.12$ .

Total area $=0.16+0.12+0.12=0.40$ . …
Start your 14-day free trial to see the full solution →
AP EAPCET 2022Set eng-2022-07-07-FN1 markMCQ
$v = e^{-\beta x}$
›Reveal solutionSolution
Separating variables in $\frac{dx}{dt}=e^{-\beta x}$ and applying $x(0)=0$ gives $x(t)=\frac1\beta\ln(1+\beta t)$ .

Concept and Intuition

This is a separable first-order ODE relating velocity to position. Rearrange so all $x$ -terms are on one side and $t$ -terms on the other, then integrate and apply the initial condition to fix the constant.

Step-by-Step Solution

$v=\dfrac{dx}{dt}=e^{-\beta x}$ .

Separate variables: $e^{\beta x}\,dx=dt$ .

Integrate both sides: $\displaystyle\int e^{\beta x}dx=\int dt \Rightarrow \dfrac{1}{\beta}e^{\beta x}=t+C$ .

Apply $x=0$ at $t=0$ : $\dfrac1\beta e^0=0+C\Rightarrow C=\dfrac1\beta$ .

So $\dfrac1\beta e^{\beta x}=t+\dfrac1\beta \Rightarrow e^{\beta x}=\beta t+1$ . …
Start your 14-day free trial to see the full solution →
AP EAPCET 2022Set eng-2022-07-08-FN1 markMCQ
$Q. The correct position$
(x) – time (t) graph for a particle moving with negative acceleration is (A) [FIGURE] (an x-t graph shaped like an inverted-U / downward-opening curve: x rises from the origin, reaches a peak, then falls back down towards the t-axis) (B) [FIGURE] (an x-t graph that is a straight line with negative slope, x decreasing linearly with time from a positive starting value) (C) [FIGURE] (an x-t graph that curves upward with increasing slope, concave up, rising steeply from the origin) (D) [FIGURE] (an x-t graph shaped like a U / upward-opening curve: x starts high, decreases to a minimum, then increases again, concave up)
›Reveal solutionSolution
Negative (constant) acceleration means the $x$ – $t$ graph must be concave DOWN; among the four described graphs, only the dome-shaped curve (A) is concave down.

Concept and Intuition

The second derivative of $x$ with respect to $t$ is the acceleration: $a = d^2x/dt^2$ . So the sign of acceleration directly controls the concavity (curvature) of the position-time graph, not its slope's sign. A negative acceleration means the graph curves downward (concave down) everywhere, like $x=v_0t-\tfrac12|a|t^2$ , a downward parabola. A straight line has zero curvature (zero acceleration) regardless of its slope.

Step-by-Step Solution

For constant acceleration $a$ (here $a<0$ ), position is $x(t) = x_0 + v_0 t + \tfrac12 a t^2$ .

Since $a<0$ , the coefficient of $t^2$ is negative, so the curve is a downward-opening parabola — concave down everywhere.

Evaluate each option:

(A) Dome/inverted-U starting at the origin, rising to a peak, then falling: this is exactly a downward parabola through the origin — concave down. ✓

(B) Straight line with negative slope: $d^2x/dt^2=0$ (zero acceleration, just negative velocity). ✗ …
Start your 14-day free trial to see the full solution →
AP EAPCET 2021Set ap-2021-09-06-FN1 markMCQ
$Q. The velocity-time graph of an object is as shown. [FIGURE] (a v-t graph with axes v vs t, showing five labelled regions in order along the t-axis: A - a rising straight line from the origin up to t1, above the axis; B - a rectangular region below the t-axis from t1 to t2; C - a rectangular region above the t-axis from t2 to t3; D - a straight line sloping below the t-axis from t3 to t4; E - a trapezoidal region above the t-axis from t4 to t5) The displacement during the interval 0 to t4 is ____ (A) area(A) + area(B) + area(C) + area(D) + area(E) (B) area(A) - area(B) + area(C) - area(D) (C) area(A) + area(B) + area(C) + area(D) (D) area(A) - area(B) + area(C) + area(D) + area(E)$
›Reveal solutionSolution
Displacement is the net area between the velocity-time graph and the time axis, counting area above the axis as positive and area below as negative. For the interval 0 to t₄, only regions A, B, C, and D lie within that time span; region E is after t₄ and does not contribute. The correct expression is area(A) − area(B) + area(C) − area(D), which matches option (B).

The key idea is simple: on a velocity-time graph, displacement equals the signed area — area above the time axis is positive (motion forward), area below is negative (motion backward). The question asks for displacement from 0 to t₄, so we only consider regions that fall within that time interval. Region E lies entirely after t₄, so it is irrelevant.

Let’s walk through it step by step.

Identify which regions lie between 0 and t₄.

The time axis is marked 0, t₁, t₂, t₃, t₄, t₅.

Region A: from 0 to t₁ (above axis) → inside interval.

Region B: from t₁ to t₂ (below axis) → inside interval.

Region C: from t₂ to t₃ (above axis) → inside interval.

Region D: from t₃ to t₄ (below axis) → inside interval.

Region E: from t₄ to t₅ (above axis) → outside the interval 0 to t₄. So only A, B, C, D matter.

Assign signs to each region.

Displacement = sum of signed areas.

Above axis → positive: A and C.

Below axis → negative: B and D. Therefore, displacement = $+A - B + C - D$ .

Check the options.

(A) includes E, which is wrong.

(B) is exactly $A - B + C - D$ — correct.

(C) treats B and D as positive, which is wrong. …
Start your 14-day free trial to see the full solution →
AP EAPCET 2021Set eng-2021-08-20-AN1 markMCQ
$s$
›Reveal solutionSolution
This tests the kinematic relation $v^2=2as$ for uniformly accelerated motion from rest, and translating it into the shape of the $s$ – $v$ graph. The answer is the upward-opening parabola through the origin, option (C).

Concept and Intuition

For 1-D motion with constant acceleration $a$ starting from rest ( $u=0$ ) at $s=0$ , the kinematic equation $v^2=u^2+2as$ reduces to $v^2=2as$ , i.e. $s=\dfrac{v^2}{2a}$ . This says $s$ is proportional to $v^2$ , not to $v$ — so on an $s$ (vertical) vs. $v$ (horizontal) graph, we get a parabola opening upward through the origin, not a straight line, and it gets steeper (not flatter) as $v$ grows.

Step-by-Step Solution

The object starts at $s=0$ with $v=0$ and undergoes constant acceleration $a$ along its direction of motion.

Kinematics: $v^2=u^2+2as=2as$ (since $u=0$ ).

Solve for $s$ : $s=\dfrac{v^2}{2a}$ , i.e. $s$ is a quadratic (parabolic) function of $v$ .

Since the coefficient $1/(2a)$ is positive, the parabola opens upward, passing through the origin, and its slope $ds/dv=v/a$ increases as $v$ increases — i.e., it curves upward more steeply, not flattening out. …
Start your 14-day free trial to see the full solution →

←Previous questionQ8 — The position–time (x versus t) graph of a particle moving in a straigh…→Next questionQ10 — A spring with one end attached to a mass and the other to a rigid supp…

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

🎓Unlock everything free for 14 days

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

Unlock everything free for 14 days