A hotel elevator ascends 200 m with a maximum speed of 5.0 m/s. Its acceleration and deceleration both have a magnitude of 1.0 m/s². How long does it take to make the complete trip from bottom to top?

Verified step by step guidance

1. Determine the time required for the elevator to accelerate to its maximum speed. Use the kinematic equation: $v\; =\; u\; +\; at$, where $v$ is the final velocity (5.0 m/s), $u$ is the initial velocity (0 m/s), and $a$ is the acceleration (1.0 m/s²). Solve for $t$.
2. Calculate the distance covered during the acceleration phase using the kinematic equation: $s\; =\; ut\; +\; \backslash (\backslash frac\{1\}\{2\}\backslash )at^2$. Substitute $u\; =\; 0$, $a\; =\; 1.0\; \backslash (\backslash text\{\; m/s\}\backslash )^2$, and the time $t$ found in the previous step.
3. Determine the distance traveled at constant speed. Subtract the distances covered during acceleration and deceleration from the total distance (200 m). The remaining distance is traveled at the maximum speed of 5.0 m/s.
4. Calculate the time spent traveling at constant speed using the formula: $t\; =\; \backslash (\backslash frac\{s\}\{v\}\backslash )$, where $s$ is the distance traveled at constant speed and $v$ is the maximum speed (5.0 m/s).
5. Add the time for acceleration, the time for deceleration (which is equal to the time for acceleration), and the time for constant speed to find the total time for the trip.

Key Concepts

Kinematics

Kinematics is the branch of mechanics that describes the motion of objects without considering the forces that cause the motion. It involves concepts such as displacement, velocity, acceleration, and time. In this problem, kinematic equations will be used to determine the time taken for the elevator to ascend 200 m, factoring in its maximum speed and acceleration.

Acceleration

Acceleration is the rate of change of velocity of an object with respect to time. It can be positive (speeding up) or negative (slowing down). In this scenario, the elevator has a constant acceleration of 1.0 m/s² during its ascent, which affects how quickly it reaches its maximum speed and how long it takes to complete the trip.

Uniform Motion

Uniform motion refers to the motion of an object moving at a constant speed in a straight line. In this case, once the elevator reaches its maximum speed of 5.0 m/s, it will travel at this constant speed until it needs to decelerate. Understanding the transition between acceleration, uniform motion, and deceleration is crucial for calculating the total time for the elevator's trip.