Question

An elevator travels up and down. The vertical position of the elevator in meters over time is shown below. Graph of vertical position x (in meters) on y axis and time (in seconds) on x axis. The initial position is 12 m at t=0 s, and the position is constant until t=8 s. Then the position decreases linearly to 0 m at t=16 s, then decreases linearly (with a sharper slope) to -15 m at t=20 s, then increases linearly to -6 m at t = 24 s. Graph of vertical position x (in meters) on y axis and time (in seconds) on x axis. The initial position is 12 m at t=0 s, and the position is constant until t=8 s. Then the position decreases linearly to 0 m at t=16 s, then decreases linearly (with a sharper slope) to -15 m at t=20 s, then increases linearly to -6 m at t = 24 s. What is the displacement of the elevator between 8 s 8 s8, start text, space, s, end text and 20 s 20 s20, start text, space, s, end text? m mstart text, m, end text What is the distance traveled by the elevator between 8 s 8 s8, start text, space, s, end text and 20 s 20 s20, start text, space, s, end text? m mstart text, m, end text

Answer 1

Ive got you deary!

To find the displacement of the elevator between 8 seconds and 20 seconds, you can calculate the change in its position during this time interval.

- At t = 8 seconds, the position is 12 meters.
- At t = 20 seconds, the position is -15 meters.

To find the displacement, you subtract the initial position from the final position:

Displacement = Final Position - Initial Position
Displacement = (-15 meters) - (12 meters) = -27 meters

So, the displacement of the elevator between 8 seconds and 20 seconds is -27 meters.

Now, to find the distance traveled by the elevator during this time interval, you can sum up the magnitudes of its position changes:

Distance = |Final Position - Initial Position|
Distance = |-15 meters - 12 meters| = |-27 meters| = 27 meters

So, the distance traveled by the elevator between 8 seconds and 20 seconds is 27 meters.