An elevator is moving upward at 0.91 m/s when it experiences an acceleration 0.31 m/s2 downward, over a distance of 0.61 m. What will its final velocity be?

Given:

u = 0.91 m/s

a=0.31m/s∧2

s = 0.61 m

s = ut + 1/2 (at∧2)

where s is the displacement of the object

u is the initial velocity

t is the time

a is the acceleration

Substituting the values

0.61=0.91*t + (0.31 * t∧2) / 2

0.61=0.91 t + 0.155 t∧2

t=0.61 secs

Consider the equation

v=u + at

where v is the initial velocity

u is the initial velocity

a is the acceleration

t is the time

Substituting the values we get

v = 0.91 + (0.31*0.61)

v = 1.099 m/s