Question

A body is placed inside a lift. If the lift falls down with an acceleration 11m/s^2 ,It's displacement in 1st second is​

Answer 1

Given:

The lift falls down with an acceleration 11m/s²

To find:

Displacement in 1st second.

Calculation:

General expression of displacement of a body in "n" th second is :

$\boxed{\sf{ \therefore \: d = u + \dfrac{1}{2} a(2n - 1)}}$

Therefore , Displacement of Body (considering it to be tightly fitted inside the lift)

$= > \: d = 0 + \bigg \{ \dfrac{1}{2} \times 11 \times (2n - 1) \bigg \}$

$= > \: d = \dfrac{1}{2} \times 11 \times (2n - 1)$

$= > \: d = \dfrac{1}{2} \times 11 \times \{(2 \times 1)- 1 \}$

$= > \: d = \dfrac{1}{2} \times 11 \times \{1 \}$

$= > d = 5.5 \: m$

So, final answer is:

$\boxed{ \bold{ d = 5.5 \: m}}$

Answer 2

Question:-

A body is placed inside a lift. If the lift falls down with an acceleration 11m/s², it's displacement in 1st second is?

Answer:-

Displacement at nth second is given by:-

$\boxed{\sf{{s}_{nth}\:=\:u\:+\:\frac{1}{2}a(2n\:-\:1)}}$

• So, here as we have to find displacement in 1st second, n = 1
• And given, a = 11m/s
• And u = 0m/s (as we have considered that the lift falls down from rest)

So, Displacement in 1st second:

$\sf{{s}_{1st}\:=\:0\:+\:\frac{1}{2}(11)[(2\:*\:1)\:-\:1}]$

$\implies\sf{{s}_{1st}\:=\:\frac{1}{2}(11)[2\:-\:1}]$

$\implies\sf{{s}_{1st}\:=\:\frac{1}{2}(11){1}}$

$\implies\sf{{s}_{1st}\:=\:5.5m}$

So Ans is

$\boxed{\bf{\red{{s}_{1st}\:=\:5.5m}}}$