Answer to Question #23311 in Mechanics | Relativity for Gabriel

Question

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?

Answer 1

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