As a descending elevator approaches the correct floor, it slows to a stop with a constant acceleration of magnitude 1m/s*2. The elevator descends 4.5m 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?

E) Δx = vt - 1/2at²

Explanation:

In order to decide which kinematic formula is most useful in finding the time taken for the elevator to stop, let's list out the relevant variables in this problem first.

Take the downwards direction to be negative and the upwards direction to be positive.

a = -1 m/s² Δy = -4.5 m v = 0 m/s (since the elevator slows to a stop) t = ?

We want an equation that relates all four of these variables.

All of the equations given in the answer choices A-D have the initial velocity in them, v₀. Since we do not have this variable, the only answer choice with acceleration, displacement, final velocity, and time is the last equation.

Δx = vt - 1/2at²

We would use this equation with respect to the y-direction.

Therefore, answer choice (E) is the correct answer. It is the most useful kinematic formula to solve for the unknown variable t.

Ithink it is letter b, i may be wrong but i think that is the answer

Initial velocity,  u = 4.9 m/sfinal velocity = v m/sinitial height,  h₁ =  245 m    final height, h₂ = 0a = -9.8  m/s²time taken = t secondss = ut + 0.5at²⇒  (h₂-h₁) = ut + 0.5at²⇒  0-245 = 4.9t + 0.5×(-9.8)×t²⇒ -245 = 4.9t - 4.9t²⇒ 4.9t² -4.9t -245 =0solving it, we get   t = 7.59sv  = u +  at = 4.9 -9.8×7.59 = 4.9 - 74.38 = -69.48 m/svelocity is  69.48 m/s downward