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

a=d/t

a="triangle"v/t

explanation:

a = acceleration

the force represented by vector a is normal force. if the hockey puck is moving to the right, when it is hit with a significant force by the hockey stick the puck will move off in a different direction. correct on edge.