Question

As a descending elevator approaches the correct floor, it slows to a stop with a constant acceleration of

magnitude 1 . The elevator descends 4.5 m while it is slowing. We want to find how many seconds the

elevator takes to slow to a stop.

Which kinematic formula would be most useful to solve for the target unknown?

Answer 1

`$y(t)=y_(0) +v_(0)t-(1)/(2)at^2 $`

Step-by-step explanation:

This is the equation of motion in general form that takes into account acceleration, initial velocity and the height above the ground and tells the position at any instance(time).

in the give problem, there is one missing piece of information that you must add, the initial velocity,
`$$v_(0) $$`.

Now you can substitute 4.5m for
`$$y_(o)$$`, 0m/s for
`$$v_(0) $$` and 1m/
`s^2` for deceleration (1 magnitude) and find that the time that is needed is 4.2426Seconds.

Bear in mind, you have set
`y(t) = -4.5m`, and just stick with negative and positive convention , up = positive, down = negative.