A fast elevator in a skyscraper is employing the following scheme to get from the starting level to the desired level: For the first half of the time it accelerates at the maximum allowed acceleration of and for the second half it decelerates at the maximum allowed acceleration of . How long does it take (in seconds) to go from ground level to the highest level at elevation?

mass and volume are the answers you are looking for

the minimum constant speed at which she can catch the train is 2.7 m/s

given:

acceleration = 0.6

time = 4.5 s

to find:

velocity = ?

formula used:

from 1st kinematic equation we get,

v = u + at

u = initial speed of train = 0 m/s

v = final speed of the train

a = acceleration of the train

t = time

solution:

from 1st kinematic equation we get,

v = u + at

u = initial speed of train = 0 m/s

v = final speed of the train

a = acceleration of the train

t = time

v = 0 + (0.6 × 4.5)

v = 2.7 m/s

the minimum constant speed at which she can catch the train is 2.7 m/s

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