The minute hand of a clock is 5.5 cm long. What is average velocity vector for the tip of the hand during the interval from the hour to 20 minutes past the hour, expressed in a coordinate system with they-axis toward noon and x-axis toward 3 o'clock? (Answer in terms of i-hat and j-hat components please)
Everyone here has helped me so much, so I will do the same. Ok you know that in 20 minutes, the minute hand moves 1/3 of the circle. Therefore it moves 120 degrees. Drawing a picture helps here. Now you have a isosceles triangle with sides 5.5 and vertex angle 120. Solve to get the other side and divide by 20 for an answer in cm/min.
Velocity is equal: v=r x ω ω is the angular velocity of the minute hand. You know that the minute hand makes 2pi in one minute (60 seconds). So angular velocity is equal: ω= 2*pi/60 = 1/60 pi. Velocity (perpendicular on the minute hand) is equal to :v=r x ω=0,055 * 1/60 pi. Velocity in i & j is v= v_{x}i + v_{y}j. v_{x}=-v*cos60 v_{y}=-v*sin60 (I have attached a photo)