A lift is coming from 8^ th floor and is just about to reach 4^ t… - Vidyadip

MCQ

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

✓
$x < 0, v < 0, a > 0.$
B
$x > 0, v < 0, a < 0.$
C
$x > 0, v < 0, a > 0.$
D
$x > 0, v > 0, a < 0.$

✓

Answer

Correct option: A.

$x < 0, v < 0, a > 0.$

Key concept: The time rate of change of velocity of an object is called acceleration of the object.
It is a vector quantity. Its direction is same as that of change in velocity $($Not of the velocity$).$
In the table: Possible ways of velocity change.

When only direction of velocity changes.	When only magnitude of velocity changes.	When both magnitude and direction of velocity change.
Acceleration perpendicular to velocity.	Acceleration parallel or antiparallel to velocity.	Acceleration has two components-one is perpendicular to velocity and another parallel or antiparallel to velocity.
E.g.: Uniform circular motion	E.g.: Motion under gravity.	E.g.: Projectile motion.

Here we will take upward direction positive. As. the lift is coming in downward direction, the displacement will be negative. We have to see whether the motion is accelerating or retarding.
We know that due to downward motion displacement will be negative. When the lift reaches $4^{th}$ floor and is about to stop velocity is decreasing with time, hence motion is retarding in nature.
Thus, $x < 0; a > 0.$
As displacementisinnegativedirection, velocity will also be negative, i.e. $v < 0.$
The motion of lift will be shown like this.

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