# Answer to Question #23311 in Mechanics | Relativity for Gabriel

Question #23311

An elevator is moving upward at 1.20 m/s when it experiences an acceleration 0.33 m/s2 downward, over a distance of 0.69 m. What will its final velocity be?

Expert's answer

The distance equal:

S = v*t - a*t^2/2

v - initial velocity

a - acceleration

t - time

t^2 - 2*v/a *t + 2*S/a = 0

t^2 - 7.2 t + 4.14 = 0

Solutions of this equation are:

t1& = 0.63 s &

t2 = 6.64 s

t1 - this is a time when elevator was moving upward

t2 - this is a time when elevator was moving downward

We need a time when elevator was moving upward, so t=t1 ->

v_f = v - a*t

v_f - final velocity

v_f = 1.2 - 0.63*0.33 = 1 m/s

Answer: final velocity will be 1 m/s

S = v*t - a*t^2/2

v - initial velocity

a - acceleration

t - time

t^2 - 2*v/a *t + 2*S/a = 0

t^2 - 7.2 t + 4.14 = 0

Solutions of this equation are:

t1& = 0.63 s &

t2 = 6.64 s

t1 - this is a time when elevator was moving upward

t2 - this is a time when elevator was moving downward

We need a time when elevator was moving upward, so t=t1 ->

v_f = v - a*t

v_f - final velocity

v_f = 1.2 - 0.63*0.33 = 1 m/s

Answer: final velocity will be 1 m/s

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