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A hotel elevator ascends 200m with maximum speed of 5m/s. Its acceleration and deceleration both have a magnitude of 1.0m/s2. Part APart complete How far does the elevator move while accelerating to full speed from rest? Express your answer with the appropriate units. 12.5 m Previous Answers Correct Part B How long does it take to make the complete trip from bottom to top? Express your answer with the appropriate units.

Sagot :

Answer:

45 s .

Explanation:

The accelerator will first accelerate , then move with uniform velocity and at last it will decelerate to rest .

displacement s = ?

acceleration a = 1 m /s²

Final speed v = 5 m/s

initial speed u = 0

v² = u² + 2as

5² = 0 + 2 x 1 x s

s = 12.5  m

B)  Let time of acceleration or deceleration be t

v = u + a t

5 = 0 + 1 t

t = 5 s

Similarly displacement during deceleration = 12.5 m

Total distance during uniform motion = 200 - ( 12.5 + 12.5 ) =  175 m .

velocity of uniform motion = 5 m /s

time during which there was uniform velocity = 175 / 5 = 35 s

Total time = 5 + 35 + 5 = 45 s .

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