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Unit 3 Review 12-1 A high-speed escalator travels at 180 ft/min. What is the absolute velocity of a person running at 700 ft/min (a) in the same direction as the escalator’s motion and (b)in a direction opposite to that of the escalator’s motion?a)VA=VB+VB/A=180ftmin+700ftmin=880ftmin⇒b)VA=VB+VBA=180ftmin−700ftmin=−520ftmin⇐12-10 a) ωB=(400revmin)(2π radrev)(min60sec)=41.89radsecvB=(rAB)(ωB)=(175mm)(1m1000mm)(41.89radsec)=7.33msecWhen θ=0, point C has no angular velocity because it is horizontally locked, so the velocity is vB=vC+vB/C=7.3304ms+0=7.3304ms⇒(positive horizontal)b) When θ=90 velocity at B is still the same as our prior calculation
vB=7.3304ms⇒sina=75∗10−3.5→a=sin−1(.15)=8.63°∡ϕ=180°−90°−8.63°=81.37°vC=vBcotϕ=7.334ms∗cot(81.37°)=1.112ms⇒12-12 To get the measure of angle B for the vector triangle here, we useα=tan−1