a)What is the velocity of the elevator at the end of the fourth 5 s interval (at
ID: 1300101 • Letter: A
Question
a)What is the velocity of the elevator at the end of the fourth 5 s interval (at 20 s)?
b)What is the displacement of the elevator above the starting point at the end of the fourth 5 s interval (at 20 s)?
where the acceleration of gravity is approximately 10 m/s^2 and the positive directions for displacement, velocity, and acceleration are upward. At time t = 0 s, an elevator is at a displacement of x= 0 m with a velocity of v= 0 m/s. A student whose normal weight is 400 N stands on a scale in an elevator and records the scale reading as a function of time. The data are shown in the graph.Explanation / Answer
0 to 5s
a = 0
x1 = 0
from 5s to 10 s
N = mg + ma1
700 = 400 + 40a1
a1 = 7.5 m/s^2
v1 = vo + a1*(10-5) = 37.5 m/s
y1 = 0.5*a*t1^1 = 0.5*7.5*25 = + 93.75 m
from 10s to 15s
N = 0
a = -g
v2 = 37.5 -g*5 = 37.5 - (10*5) = -12.5 m/s
y2 = y1 + v1*5-0.5*g*t2^2 = 93.75 + (37.5*5) - (0.5*10*25) = 156.25 m
15s to 20s
N = mg + ma3
600 = 400 + 40*a3
a3 = 5 m/s^2
V3 = -12.5+ (5*5) = 12.5 m/s <------answer
y3 = y2 + v2*(20-15) + (0.5*a*5^2
y3 = 156.25-(12.5*5)+(0.5*5*5*5)
y3 = 156.25 m <-------answer
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