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natta225 [31]
3 years ago
12

An elevator accelerates uniformly from rest to a speed of 2.5 m/s in 12s. What is the distance the elevator travels during this

time? must show work.
Physics
1 answer:
Aleks04 [339]3 years ago
7 0

We use the kinematic equation,

s= u t +\frac{1}{2}a t^2                      

Here, s is the distance traveled by elevator u is the initial velocity and t is time.

As elevator accelerates uniformly from rest to a speed of 2.5 m/s in 12 s, so

a = \frac{v_{f}- v_{i}  }{t} = \frac{2.5 m/s-0}{12 s} = 0.208 \ m/s^2

Thus, the distance the elevator travels during 12 s time is

s= 0+\frac{1}{2} 0.208 \ m/s^2(12)^2 = 14.97 m \simeq  15 m


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