An elevator goes up twenty feet in five seconds. Use approximations in this prob

Question

An elevator goes up twenty feet in five seconds. Use approximations in this problem. a. How fast does the elevator go? b. It takes half a second to reach its maximum speed, and another half-sccond to decelerate to rest. Calculate the acceleration for both cases. c. Calculate the distance for both cases. d. How do parts b and c change part a's answer? This is an example of an iterative procedure, approximating an answer, when an solution is complicated or non-existant. One begins with a reasonable approximation, then calculates a correction. One may repe answer. at the process, and the answer converges to the exact Extra Credit: solve the problem exactly for the constant speed, the acceleration, and the acceleration distance.

Explanation / Answer

(a) We have to neglect the acceleration due to gravity and assume that the velocity is constant.
Average speed= distance/time
= 20(ft)/5 (s) = 4ft/s
hence the elevator is moving with a speed of 4 ft/s
(b) initial speed of elevator (u) = 0 (starting from rest)
Final speed of elevator (V) = 4 ft/s
Time taken = 0.5 s
from kinematic equation
V = u +at
4 = 0 +a*0.5
a = 8 ft/s2
hence the acceleration of the elevator is 8 ft/s2
now for the deacceleration
Final speed of the elevator = 0
Initial speed of the elevator (u) = 4 ft/s
Time taken (t) = 0.5 s
V = u +at
0 = 4+0.5a
a = -8 ft/s2
-ve sign shows that the acceleration is opposite to that of motion means it is deacceleration
(c) Now for the accelerating condition
S = ut+(1/2)at2
S = 0 + (1/2)*8*(0.5)2 = 1 m
hence the elevator will cover 1 m while accelerating
Now distance cover while deacceleration
S = ut +(1/2)at2
S = 4*0.5 +(1/2)*(-8)*(0.5)2 = 1 m
hence in the deacceleration also elevator will cover 1 m
(d) It will not affect it by changing part b and C .

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