An elevator ascends to the top of a building. 200 meters, with a maximum speed o

Question

An elevator ascends to the top of a building. 200 meters, with a maximum speed of 5.0 m/s. Its acceleration and deceleration both have 3 magnitude of 1.0 m/s^2 How far does the elevator travel while accelerating to maximum speed from rest? How long does it take the elevator to make a complete trip from bottom of the building to the top? Provide a motion diagram next to the pictorial representation shown Please, include velocity and acceleration vectors (arrows). Please, refer to example problems in Chapter 1 for support.

Explanation / Answer

(a) Let's split elevator's journey in 3 parts -

1) Constant acceleration, a

2) Constant velocity, v

3) Constant deceleration ,- a

Let 's' be the distance travelled by elevator before it reaches a maximum speed of 5 m/s (v). Then,

v2 - u2 = 2as gives,

52 - 0 = 2 x 1 x s

or, s = 12.5 m

(b) Time for complete trip = t1 + t2 + t3

t1 (constant acceleration) = (v-u) / a = (5-0)m/s / 1 m/s2 = 5 s

t3 (constant acceleration) = t1 (constant acceleration) = 5 s

Distance travelled in each of the identical acceleration and deceleration phases is (12.5+12.5) = 25 m.

Remaining distance = (200 - 25) m = 175 m

This distance must have been covered in time t2 with a constant speed of 5 m/s. Therefore,

t2 = 175 m / 5 m/s = 35 s

Thus, total time taken for bottom -> top trip is:

5 s + 5 s + 35 s = 45 s

(c) For this part, please post link to Chapter 1 problems that have been referred to which can be used in the pictorial representation.

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