Question

) a hotel elevator ascends 200m with maximum speed of 5 m/s. its acceleration and deceleration both have a magnitude of 1.0m/s2.

Answer 1

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

Further explanation

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

Formula used

$\large{\boxed{\bold{S\:=\:v\:\times\:t}}}$

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

$\large{\boxed{\bold{St\:=\:vot\:+\frac{1}{2}at^2}}}$

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

Answer 2

a) 12.5 m

Explanation:

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

$v^2 - u^2 = 2ad$

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:

$d=\frac{v^2-u^2}{2a}=\frac{(5 m/s)^2-0}{2(1.0 m/s^2)}=12.5 m$

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

$d'=200 m- 12.5 m -12.5 m=175 m$

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

$t'=\frac{d'}{v}=\frac{175 m}{5 m/s}=35 s$

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

$t''=\frac{v-u}{a}=\frac{5 m/s-0}{1 m/s^2}=5 s$

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

$t=t'+t''+t''=35 s+5 s+5s =45 s$