) A body travels 25% of the circumference of a circular path having radius “R” cmb) A body is thrown vertically upward from top of a building which is 30 m tall. The body goes 5 m high above the top of the building and then falls TOWARDS the ground and finally HITS the ground.​To find:Find distance and displacement of a body in each of the above mentioned situations.Solution:a) Circumference of the circle = 2πr25% of the circumference of a circular path = 25% × 2πr= 25/100 × 2πr= 1/4 × 2πr= πr/2∴ The distance covered is πr/2 cm.As displacement is the shortest distance covered, so we have,= √ (r² + r²)    (using Pythagoras THEOREM)= √(2r²)= √2r∴ The displacement is √2r cm.b)A body is thrown vertically upward from top of a building which is 30 m tall. The body goes 5 m high above the top of the building.⇒ distance = 30 + 5 = 35 m.∴ The distance is 35 m and then falls towards the ground and finally hits the ground.​As the object hits the ground, the displacement is 0.∴ The displacement is 0 m.