Question

As a descending elevator approaches the correct floor, it slows to a stop with a constant acceleration of magnitude 1 ms21\,\dfrac{\text{m}}{\text{s}^2} 1 s 2 m ​ 1, start fraction, start text, m, end text, divided by, start text, s, end text, squared, end fraction . The elevator descends 4.5 m4.5\,\text{m} 4.5m 4, point, 5, start text, m, end text while it is slowing. We want to find how many seconds the elevator takes to slow to a stop.

Answer 1

KINEMATIC FORMULA

deltaX=vt-1/2at^2

Step-by-step explanation:

Answer 2

Step-by-step explanation:

Given that,

The elevator slow to stop, this shows that it is decelerating and the final velocity is 0

v = 0m/s

Constant deceleration

a = -1m/s²

It is negative because it is deceleration

The distance the elevator descend is 4.5m

S=4.5m

Then, we want to find the time the elevator spent before stopping

Using the equation of motion

v = u + at

v² = u² + 2as

Where

v is final velocity

u Is initial velocity

a is the deceleration

s- is distance traveled

From here we can find the initial velocity of the elevator

v² = u² + 2as

0² = u² - 2 × 1 × 4.5

0 = u² - 9

u² = 9

u = √9

u = 3m/s

The initial speed is 3m/s

Then, to find the time taken, we can use the first equation

v = u + at

0 = 3 - 1 × t

0 = 3 - t

t = 3 seconds

The time taken before the elevator stop is 3 secs