Question

A lift coming down is just about to reach the ground floor. Taking the ground floor as origin and positive direction upwards for all quantities, which of the following is correct $\left( {x = displacement,v = velocity,a = acceleration} \right)$
A. $x < 0,v < 0,a > 0$
B. $x > 0,v < 0,a < 0$
C. $x > 0,v < 0,a > 0$
D. $x > 0,v > 0,a > 0$

Answer 1

Hint:To answer this question, you need to understand the concept of uniformly accelerated motion.
In a uniformly accelerated motion, the body is moving under constant acceleration. When the acceleration is constant, there are 3 formulas that can be applied:
1. $v = u + at$
2. $s = ut + \dfrac{1}{2}a{t^2}$
3. ${v^2} - {u^2} = 2as$
where v = final velocity, u = initial velocity, s = displacement, t = time.

Complete step-by-step answer:
Consider the lift moving downwards with a uniformly accelerated motion and there is an additional acceleration acting on the lift due to the acceleration due to gravity, $g = 9.8m/{s^2}$ as the lift is moving downwards towards the Earth.
Let us consider the variables displacement, velocity and acceleration in these cases:

1. Displacement :
The lift is moving downwards against the upward direction as measured from the origin. Hence, the displacement is negative .
$x < 0$

2. Velocity:
Velocity, $v = \dfrac{d}{t}$
Since the displacement is in the negative direction, the rate of change of displacement with time i.e velocity is also, negative (here, it is confirmed that the velocity is negative if the displacement is negative since the velocity can be positive with negative displacement only, if the time is negative which is IMPOSSIBLE).
$v < 0$

3. Acceleration:
The acceleration due to gravity is acting downwards on the lift. Even though the body is moving downwards, the acceleration of the lift is increasing positively due to the effect of gravity. The sign of acceleration, here is irrespective of the signs of the velocity and displacement.
Hence, $a > 0$
Therefore,
Displacement, $x < 0$
Velocity, $v < 0$
Acceleration, $a > 0$

Hence, the correct option is Option A.

Note: The students sometimes assume that the acceleration on the lift is equal to the acceleration due to gravity, but IT IS NOT!. The acceleration will only be equal to g , if and only if, it is a free fall which means it is unassisted such as when the chords of the lift is being cut and there is nothing to hold the lift. But in this case, the lift is not under free fall. Here, the acceleration will be equal to $\left( {a + g} \right)$.