a) 12.5 m
Explanation:
During the acceleration phase, the distance travelled by the elevator is given by:
[tex]v^2 - u^2 = 2ad[/tex]
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:
[tex]d=\frac{v^2-u^2}{2a}=\frac{(5 m/s)^2-0}{2(1.0 m/s^2)}=12.5 m[/tex]
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
[tex]d'=200 m- 12.5 m -12.5 m=175 m[/tex]
This distance is covered at a speed of v=5 m/s, so the time taken is
[tex]t'=\frac{d'}{v}=\frac{175 m}{5 m/s}=35 s[/tex]
The time the elevator takes to cover the 12.5 m in the acceleration phase is
[tex]t''=\frac{v-u}{a}=\frac{5 m/s-0}{1 m/s^2}=5 s[/tex]
and the same for the deceleration phase; therefore, the total time for the trip is
[tex]t=t'+t''+t''=35 s+5 s+5s =45 s[/tex]
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
[tex]\large{\boxed{\bold{S\:=\:v\:\times\:t}}}[/tex]
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
[tex]\large{\boxed{\bold{St\:=\:vot\:+\frac{1}{2}at^2}}}[/tex]
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
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Keywords: Regular straight motion, Straight motion changes regularly, elevator
