)a hotel elevator ascends 200m with maximum speed of 5 m/s. its acceleration and deceleration both have a magnitude of 1.0m/s2. a. how far does the elevator move while accelerating to full speed from rest? b. how long does it take to make the complete trip from bottom to top

a) 12.5 m

Explanation:

During the acceleration phase, the distance travelled by the elevator is given by:

where

v = 5 m/s is the final speed of the elevator

u = 0 is the initial speed

a = 1.0 m/s^2 is the acceleration

d is the distance travelled

Using the equation, we can find d:

b) 45 s

Explanation:

The distance covered during the acceleration phase is 12.5 m. The distance covered during the de-celeration phase, at the end of the trip, is exactly the same (because the values of u, v and a are the same). The total distance is 200 m, so the distance covered at a constant speed of 5 m/s is

This distance is covered at a speed of v=5 m/s, so the time taken is

The time the elevator takes to cover the 12.5 m in the acceleration phase is

and the same for the deceleration phase; therefore, the total time for the trip is

The distance of the elevator at the start while accelerating to full speed from rest = 12.5 m

the complete trip from bottom to top takes time: 45 s

Regular straight motion is the motion of objects on a straight track that have a fixed speed

Formula used

S = distance = m

v = speed = m / s

t = time = seconds

Straight motion changes regularly is the straight motion of objects that have a fixed acceleration

Formula used

V = vo + at

Vt² = vo² + 2as

St = distance on t

vo = initial speed

vt = speed on t

a = acceleration

In the motion of the elevator occurs two movements, namely straight regular motion and straight motion change regularly

Regular straight motion occurs when the elevator is running at a steady speed and regular motion changes regularly when the elevator starts moving and when it wants to stop

Known

a = acceleration = deceleration = 1 m/s²

So the distance of the elevator at the start and want to stop is the same

conditions will move

Vt² = vo² + 2as

Vo = 0 = silent condition

Vt = 5 m / s

a = 1

5² = 0 + 2as

25 = 2as

25 = 2.1.s

s = 25 : 2

s = 12.5 m

Time taken:

Vt = vo + at

5 = 0 + 1.t

t = 5 s

the condition when it wants to stop (deceleration = a negative value)

Vt = 0 (stop)

Vo = 5

0  = 5² -2.1.s

2s = 25

s = 12.5 m

Time taken:

Vt = vo - at

0 = 5 - 1.t

t = 5 s

Because the total height = 200 m, the remaining distance (which has a fixed speed) is:

200 m - 12.5 -12.5 = 175 m

Because this is a straight-line motion with a fixed speed, the time taken:

S = v x t

t = s: v

t = 175: 5

t = 35 s

So the total time taken:

5 s + 35 s + 5 s = 45 s

Classify the motion of this object.

link

three different examples of accelerated motion

link

Describe the direction, speed, and acceleration of an object moving with uniform circular motion

link

object speeds up, slows down, or changes direction

link

Keywords: Regular straight motion, Straight motion changes regularly, elevator