f.4,,phys,,mechanics,,about lift.

2008-09-26 5:05 am
A lift can accelerate or decerate uniformly at 1.5ms^-2. It accelerates for 1s when starting from rest, and decelerates for 1s before arriving at the destination. It travels at a constant velocity in between.

a) What is the maximum speed of the lift?
b) How long would a person take to travel from the ground floor to the fifth floor by the lifr, assuming that rach door has a height of 3m?

回答 (1)

2008-09-26 5:14 am
✔ 最佳答案
a. By v = u + at

v = 0 + (1.5)(1)

Max. speed, v = 1.5 ms-1


b. Consider the journey when accelerating.

By v2 = u2 + 2as

(1.5)2 = 0 + 2(1.5)s

Distance travelled when accelerating, s = 0.75 m

The distance travelled when decelerating is also = 0.75 m

Therefore, the distance travelled when the motion is uniform, s

= 3 X 5 - 0.75 - 0.75

= 13.5 m

Time required to travel the distance s under uniform speed of 1.5 ms-1

= 13.5 / 1.5 = 9 s

Therefore, the total required time

= 1 + 1 + 9

= 11 s



2008-09-26 08:11:02 補充:
因為b part的運動可分為三段,1. 加速,2. 均速,3. 減速。所以要分三part去計算。
參考: Myself~~~


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