Question

a long hand of a clock measures 28 cm. find the distance & displacement of the long hand from 4.00 pm to 5.30 pm .

Answer 1

Let the length of long hand be the radius of the clock 
first find the cicumference of the clock by the formula 
C = 2 pi r 
C = 2 x 22/7 x 28
C = 2 x 22 x 4
C = 44 x 4
C = 176

now make 12 divisions of the received radius by dividing 176/12

one div = 14.6
for 4 pm to 5:30
= 14.6 x 1.5
= distance= 7.3 cm 
= displacement = 0 cm

Answer 2

see the diagram.

     the long hand AB of the clock has two ends.  One end A is fixed at the center of the circle in which the other end B of the hand rotates.

   The radius R of the circle in which the tip of the long hand moves = length of the hand.
            R = 28 cm.

At 4 pm and at 5:30 pm the long minutes hand of the clock is in positions AB and AB'.  BAB' is a straight line.
 
  The long hand (minutes hand) moves 360 deg. in 1 hour.  So in one and half hours, it rotates  360° * 1.5 = 540°  or  3 π radians.

  The angular displacement θ of the clock hand
            = angle BAB' = 180 degrees = π radians

 The total distance traveled by the tip of the hand B
     = 1.50 * perimeter of the circle    (as B makes 1 complete revolution and 1/2 revolution)
     = 1.50 * 2 π R
     = 3 * 22/7 * 28 cm
     = 264 cm

  The net displacement traveled by the tip of the hand B
     = 1/2 * perimeter of the circle        (as B moves from 0 to 6).
     = 1/2 * 2 π R
     = 22/ 7 * 28 cm
     = 88 cm