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C = tau*r = 5m*tau
v = d/t & d = vt = 0.5m/s*5s = 2.5m (d is the displacement)
P = 2.5m/(5m*tau) (P is the proportion moved around the circle)
theta = P*tau = 1/2rad (theta is the angle moved through)

Then form an isosceles triangle with the 'base' as the line between the two points. The problem is now: we have two 5m lines whose ends join at a point at an angle of 1/2rad to eachother. Calculate the end of the second line given the equation of the first line.