Question

A certain elevator cab has a total run of 218 m and a maximum speed is 319 m/min, and it accelerates from rest and then back to rest at 1.20 m/s^2. (a) How far does the cab move while accelerating to full speed from rest? (b) How long does it take to make the nonstop 218 m run, starting and ending at rest?

Answer 1

Answer:

a)11.6m

b)45.55s

Explanation:

A body that moves with constant acceleration means that it moves in "a uniformly accelerated movement", which means that if the velocity is plotted with respect to time we will find a line and its slope will be the value of the acceleration, it determines how much it changes the speed with respect to time.

When performing a mathematical demonstration, it is found that the equations that define this movement are as follows.

Vf=Vo+a.t  (1)\\\\

{Vf^{2}-Vo^2}/{2.a} =X(2)\\\\

X=Xo+ VoT+0.5at^{2}    (3)\\

X=(Vf+Vo)T/2 (4)

Where

Vf = final speed

Vo = Initial speed

T = time

A = acceleration

X = displacement

In conclusion to solve any problem related to a body that moves with constant acceleration we use the 3 above equations and use algebra to solve

a)

for this problem

Vo=0

Vf=319m/min=5.3m/s

a=1.2m/s^2

we can use the ecuation number 1 to calculate the time

t=(Vf-Vo)/a

t=(5.3-0)/1.2=4.4s

then we use the ecuation number 3 to calculate the distance

X=0.5at^2

X=0.5x1.2x4.4^2=11.6m

b)second part

We know that when the elevator starts to accelerate and decelerate, it takes a distance of 11.6m and a time of 4.4s, which means that if the distance is subtracted 2 times this distance (once for acceleration and once for deceleration)

we will have the distance traveled in with constant speed.

With this information we will find the time, and then we will add it with the time it takes for the elevator to accelerate and decelerate

X=218-11.6x2=194.8m

X=VT

T=X/v

t=194.8/5.3=36.75s

Total time=36.75+2x4.4=45.55s