We are asked to find the time it takes for the elevator to reach a height of 80 m, given that it accelerates for the first half of the time with an acceleration of $1.0 \, \text{m/s}^2$, and then decelerates at $-1.0 \, \text{m/s}^2$ for the second half.

Step-by-step solution:

Let:

• $a = 1.0 \, \text{m/s}^2$ (acceleration during the first half),
• $-a = -1.0 \, \text{m/s}^2$ (deceleration during the second half),
• $d = 80 \, \text{m}$ (total displacement),
• $t$ be the total time taken.

1. Time Symmetry

Because the elevator accelerates and decelerates symmetrically, the time taken to accelerate is the same as the time taken to decelerate. Let's denote the time to accelerate (first half) as $t_1 = \frac{t}{2}$.

2. Displacement During Acceleration

For the first half of the motion (accelerating phase), we use the equation of motion: $d_1 = \frac{1}{2} a t_1^2$ where $d_1$ is the displacement during the acceleration phase.

Since the motion is symmetric, the displacement during the deceleration phase is the same: $d_2 = d_1$ Thus, the total displacement $d = d_1 + d_2 = 2 d_1$.

3. Total Displacement

From the previous equation: $d = 2 \times \frac{1}{2} a t_1^2 = a t_1^2$ Substituting $t_1 = \frac{t}{2}$: $d = a \left( \frac{t}{2} \right)^2 = \frac{a t^2}{4}$ Given that $d = 80 \, \text{m}$ and $a = 1.0 \, \text{m/s}^2$, we can now solve for $t$.

$80 = \frac{1.0 \times t^2}{4}$ $t^2 = 320$ $t = \sqrt{320} \approx 17.89 \, \text{seconds}$

Thus, the total time it takes for the elevator to reach the highest level is approximately 17.89 seconds.

Would you like more details or have any questions?

Here are some related questions:

1. What would happen if the acceleration were doubled?
2. How would the time change if the total height were 160 meters instead of 80 meters?
3. How do the kinematic equations change if the elevator doesn't decelerate at the same rate?
4. What is the maximum velocity reached by the elevator during the ascent?
5. Can we calculate the work done by the elevator's motor over the entire journey?

Tip: When working with symmetric motion problems, breaking the motion into acceleration and deceleration phases simplifies calculations greatly.