An elevator is moving upward at 1.20 m/s when it experiences an acceleration of 0.31 downward over a distance of 0.75m. what will its final velocity be?

The final velocity is 0.99 m/s.

Explanation:

Given that,

Speed = 1.20 m/s

Acceleration = -0.31 m/s²

Distance = 0.75 m

We need to calculate the final velocity

Using equation of motion

Where, v = final velocity

u = initial velocity

a = acceleration

s = distance

Put the value in the equation

Hence, The final velocity is 0.99 m/s.

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explanation:

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