A local elevator moves upward at a constant 2.1 mps passing a stopped express elevator. Precisely 2.7 seconds later the express elevator catches up with the local elevator where the velocity of the local with respect to the express elevator is - 9.5 m/s. Determine (a) the acceleration of the express elevator in m/s/s he distance traveled in m for the

Question

Lilah Vega

Physics

6 December, 19:12

A local elevator moves upward at a constant 2.1 mps passing a stopped express elevator. Precisely 2.7 seconds later the express elevator catches up with the local elevator where the velocity of the local with respect to the express elevator is - 9.5 m/s. Determine (a) the acceleration of the express elevator in m/s/s he distance traveled in m for the

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Answers (1)

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Tucker Coleman · Answer 1

4.3 m/s² (approx)

Explanation:

Velocity of local elevator with respect to express elevator initially is 2.1 mps

After 2.7 seconds, the gap between them is

2.7 x 2.1 = 5.67 m

The express elevator begins with acceleration. At the time of catching up

velocity of local with respect to express is - 9.5 m / s. That means velocity of

express with respect to local will be 9.5 m/s.

Relative velocity of express is 9.5 m/s

Absolute velocity of express is 9.5 + 2.1 = 11.6 m/s

Acceleration of express = change in velocity / time

= (11.6 - 0) / 2.7 = 4.3 m/s² (approx)