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
<h3><em>Further explanation
</em></h3>
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
<h3><em>Learn more
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Keywords: Regular straight motion, Straight motion changes regularly, elevator