(II) Suppose the force FT in the cord hanging from thepulley of Example 10-9, Fig. 10-21, is given by the relationFT = 3.001 0.2012 (newtons) where t is in seconds. If thepulley starts from rest, what is the linear speed of a point onits rim 8.0 s later? Ignore friction.
Lectures 1 2 and 3 Average Speed & Average Velocity vavgdistancetime=dt *speed is a scalar (magnitude with no direction vavgdisplacementtime= rt * velocity is a vector (has direction) → UNITS: ms vavg and r point in direction of motion vx= xf-x0t → xf=x0+t(vx) *with constant velocity Negative slope = negative velocity Steeper slope = faster speed Acceleration aavgvt *rate of change of velocity (slowing, speeding, changing direction) aavgv - n+1 n → UNITS: ms 2 → If v and a are point in the same direction, the object is speeding up afree fall0 ms2 → 1 dimensional acceleration down a slope: as= g(sin) Ki
