Question

An elevator is moving at 1.1 m/s as it approaches its destination floor from below. When the elevator is a distance h from its destination, it accelerates with
a = −0.77 m/s2,
where the negative sign indicates a downward vertical direction. ("Up" is positive.) Find h.

Answer 1

An elevator is moving at $1.1 \, \text{m/s}$ as it approaches its destination floor from below. When the elevator is a distance $h$ from its destination, it accelerates with $a = -0.77 \, \text{m/s}^2$, where the negative sign indicates a downward vertical direction. ("Up" is positive.) Find $h$.

Solution:

Let:

$$v = 1.1 \, \frac{m}{s}$$ $$a = -0.77 \, \frac{m}{s^2}$$

$h - ?$

$$h = \frac{1}{2} a t^2$$ $$v = a t, t = \frac{v}{a}$$ $$h = \frac{1}{2} \frac{v^2}{a}$$ $$h = \frac{1}{2} * \frac{1.1^2}{0.77} = 0.79 \, \text{m}$$

Answer: $0.79 \, \text{m}$