Focal length of a convex mirror, $$f = {R \over 2} = {{20} \over 2}$$ m = 10 m

For first object, $${v_1} = + {{25} \over 3}$$ m, $$f = + 10$$ m

Using mirror formula $${1 \over v} + {1 \over u} = {1 \over f}$$

$$\therefore$$ $${1 \over {(25/3)}} + {1 \over {{u_1}}} = {1 \over {10}}$$ or $${1 \over {{u_1}}} = {1 \over {10}} - {3 \over {25}}$$

or $${u_1} = - 50$$ m

For second object,

$${v_2} = + {{50} \over 7}$$ m, $$f = + 10$$ m

$$\therefore$$ $${1 \over {{v_2}}} + {1 \over {{u_2}}} = {1 \over f}$$

$${1 \over {(50/7)}} + {1 \over {{u_2}}} = {1 \over {10}}$$ or $${1 \over {{u_2}}} = {1 \over {10}} - {7 \over {50}}$$ or $${u_2} = - 25$$ m

Speed of the object $$ = {{25} \over {30}}$$ m s^$$-$$1

$$ = {{25} \over {30}} \times {{18} \over 5}$$ km h^$$-$$1 = 3 km h^$$-$$1