Answer to Question #146181 in Mechanics | Relativity for Emma

Question

A person has a mass of 60.0 kg. She is standing on a scale in an elevator that is at rest. The elevator starts accelerating upward at a rate of 2.10 m/sec2.
(a)How far has the elevator moved upward at t = 3.15 seconds?
(b)What is the speed of the elevator at t = 3.15 seconds?

Answer 1

Given,

Mass of person=$60kg$

Accelaration of elevator $a=2.10m/s^2$

Initialy It is at rest so $u=0m/s$

a) at $t=3.15s$

Apply second equation of motion,

$s=ut+\dfrac{1}{2}at^2 \\\Rightarrow s=\dfrac{1}{2}(2.10)(3.15)^2\\\Rightarrow s=10.5(9.925)=104.18m$

Hence Distance covered by lift is$104.18m$

(b)Apply first equation of motion

Let the speed at $3.15s$ be $v$

$v=u+at \\\Rightarrow v=0+(2.10)(3.15)=6.615m/s$

Hence the velocity is $6.615m/s$