Question

A hotel elevator ascends 160 m with a maximum speed of 6.0 m/s . Its acceleration and deceleration both have a magnitude of 1.5 m/s2 .

Part A: How far does the elevator move while accelerating to full speed from rest?

Part B: How long does it take to make the complete trip from bottom to top?

Answer 1

(A)

Given that,

v = 6 m/s

u = 0

a = 1.5 m/s^2

From kinematic equation,

v^2 - u^2 = 2as

6^2 - 0 = 2*1.5*s

s = 12 m

(B)

time taken while acceletating,

v - u = at

6 - 0 = 1.5 *t

t1 = 4 s

time taken while decelerating,

t2 = 4 s

distance travelled while accelearting,

s = ut + (1/2)at^2

s = 0 + (1/2)*1.5*4^2 = 12 m

distance traveled while decelerating,

s' = 12 m

total distance = 160 - 12 - 12 = 136 m

time taken at top when speed is constant,

t3 = total distance / speed = 136 / 6

t3 = 22.6 s

total time taken to make the complete trip from bottom to top,

t = t1 + t2 + t3

t = 4 + 4 + 22.6

t = 30.6 s