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