A lift is coming from 8th floor and is just about to reach 4th floor. Taking ground floor as origin and positi

NCERT Exemplar · Q2

Q. A lift is coming from 8th floor and is just about to reach 4th floor. Taking ground floor as origin and positive direction upwards for all quantities, which one of the following is correct?

(a)

x < 0

v < 0

a > 0

(b)

x > 0

v < 0

a < 0

(c)

x > 0

v < 0

a > 0

(d)

x > 0

v > 0

a < 0

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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

A ball is dropped from rest from a height of $20\ \text{m}$ . Taking downward as positive, $u = 0$ , $a = +g = 10\ \text{m/s}^2$ , $s = +20\ \text{m}$ .

Using $s = ut + \frac{1}{2}at^2$ :

20 = 0 + \frac{1}{2}(10)t^2

20 = 5t^2

t^2 = 4 \implies t = 2\ \text{s}

The ball takes 2 seconds to hit the ground. Notice: we only needed one coordinate (the vertical $y$ -axis) to describe the entire motion.

Why This Matters

One dimensional motion is the foundation for everything else in kinematics. Once you master signs, displacement vs distance, and the constant-acceleration equations here, you can extend the same ideas to two and three dimensions by treating each dimension independently. Every projectile motion problem, for instance, is just two independent 1D motions (horizontal and vertical) happening simultaneously.

Start with the straight line. Get the signs right. The rest follows.

One Dimensional Motion is the subject of the entire NCERT Class 11 Physics chapter on Motion in a Straight Line, matching searches like "one dimensional motion: definition and equations" or "kinematics important questions class 11 physics". This chapter's signed quantities and constant-acceleration equations are foundational for nearly every mechanics question tested in CBSE boards, and remain an important topic for JEE Main and NEET.

The lift is above the ground ( $x > 0$ ), moving downwards ( $v < 0$ ), and slowing down as it approaches the 4th floor, meaning its acceleration is directed upwards ( $a > 0$ ). The correct option is (C).

When analyzing motion, it's crucial to first establish a clear coordinate system: an origin (the zero point for position), and a positive direction. The signs of position, velocity, and acceleration are entirely dependent on this chosen system.

Position ( $x$ ) tells us where an object is relative to the origin. If the object is in the positive direction from the origin, $x$ is positive. If it's in the negative direction, $x$ is negative.
Velocity ( $v$ ) tells us how fast an object is moving and in which direction. If an object moves in the positive direction, $v$ is positive. If it moves in the negative direction, $v$ is negative.
Acceleration ( $a$ ) tells us how the velocity is changing. It's the rate of change of velocity.
- If an object is speeding up in the positive direction, $a$ is positive.
- If an object is slowing down in the positive direction, $a$ is negative.
- If an object is speeding up in the negative direction, $a$ is negative.
- If an object is slowing down in the negative direction, $a$ is positive. Essentially, if acceleration is in the same direction as velocity, the object speeds up. If acceleration is in the opposite direction to velocity, the object slows down.

Let's apply these concepts to the lift's motion.

Define the Coordinate System:

The problem states: "Taking ground floor as origin and positive direction upwards for all quantities."
- Origin: Ground floor ( $x=0$ ).
- Positive direction: Upwards.
- Negative direction: Downwards.
Determine the sign of Position ( $x$ ):

The lift is "just about to reach 4th floor." The 4th floor is above the ground floor. Since the ground floor is the origin and upwards is the positive direction, any point above the ground floor will have a positive position.

Therefore, $x > 0$ .
Determine the sign of Velocity ( $v$ ):

The lift is "coming from 8th floor and is just about to reach 4th floor." This means the lift is moving from a higher floor to a lower floor. In our coordinate system, this is a downward motion. Since upwards is positive, downward motion corresponds to a negative velocity.

Therefore, $v < 0$ .
Determine the sign of Acceleration ( $a$ ):

The phrase "just about to reach 4th floor" implies that the lift is slowing down as it approaches its destination.
- The lift is moving downwards, so its velocity is negative ( $v < 0$ ).
- For an object moving in the negative direction to slow down, its acceleration must be directed opposite to its velocity, i.e., in the positive direction.
- Since upwards is the positive direction, the acceleration must be positive. Therefore, $a > 0$ .
Watch out

A common misconception is to assume that if an object is moving downwards, its acceleration must be negative. This is only true if it's speeding up downwards. If it's slowing down while moving downwards, its acceleration must be upwards (positive) to oppose the downward motion.
Combine the results:

We have found:
- $x > 0$
- $v < 0$
- $a > 0$
Comparing this with the given options:

(A) $x < 0$ , $v < 0$ , $a > 0$

(B) $x > 0$ , $v < 0$ , $a < 0$

(C) $x > 0$ , $v < 0$ , $a > 0$

(D) $x > 0$ , $v > 0$ , $a < 0$

Our findings match option (C).

✓Final answer

\boxed{\text{(C)}}

Concept: Confirming a Sign Argument with an Explicit Numeric Kinematic Model

Method: Assign Real Numbers and Compute (rather than reasoning about signs in words alone)

The verbal argument ("moving down, slowing down, so acceleration is up") is correct, but it's worth confirming with one concrete, self-consistent numeric example — assigning real floor heights, times, and speeds, and checking that a genuine kinematic model reproduces exactly the claimed signs.

Steps

Set up a coordinate system and real numbers. Ground floor $=0$ , positive direction upward. Take each floor as $3\ \text{m}$ apart, so the 4th floor is at $x = 4\times 3 = 12\ \text{m}$ and the 8th floor at $x=24\ \text{m}$ .
Model the lift's descent as decelerating (slowing down as it approaches the 4th floor to stop or ease in). Suppose it passes some point above the 4th floor moving down at $4\ \text{m s}^{-1}$ , and is down to $1\ \text{m s}^{-1}$ by the time it is "just about to reach" the 4th floor, over an interval of $\Delta t = 1\ \text{s}$ . With downward taken as negative:

v_{\text{start}} = -4\ \text{m s}^{-1}, \qquad v_{\text{end}} = -1\ \text{m s}^{-1}

Compute the acceleration from these two numbers:

a = \frac{v_{\text{end}} - v_{\text{start}}}{\Delta t} = \frac{(-1) - (-4)}{1} = +3\ \text{m s}^{-2}

Positive — i.e. directed upward — even though the lift is still moving downward the whole time. This numerically confirms the general rule: decelerating while moving in the negative direction requires a positive (upward) acceleration.

Check position and velocity at the "just about to reach 4th floor" instant. The lift is still above the ground (at or just above $x=12\ \text{m} > 0$ ), so $x>0$ ; it is still moving downward, so $v = -1\ \text{m s}^{-1} < 0$ .
Collect the three signs from the explicit model: $x>0$ , $v<0$ , $a>0$ — matching option (c) exactly, now backed by real numbers rather than only a verbal deceleration argument.

Why building the numeric model is useful

A purely verbal sign argument can occasionally hide an error (e.g. mixing up which direction is "slowing down" relative to which is "speeding up"). Plugging in an explicit, self-consistent set of numbers and computing $a=\Delta v/\Delta t$ directly removes any ambiguity — and the fact that any consistent choice of numbers reproduces the same three signs confirms the sign pattern is a structural feature of "moving down while decelerating," not an artifact of a particular number choice.

Final Answer

\boxed{x>0,\ v<0,\ a>0 \;\Rightarrow\; \text{option (c)}}

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Q.A lift is coming from 8th floor and is just about to reach 4th floor. Taking ground floor as origin and positive direction upwards for all quantities, which one of the following is correct?

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