Velocity (v) and acceleration (a) in two systems of units 1 and 2 are related as $${v_2} = {n \over {{m^2}}}{v_1}$$ and $${a_2} = {{{a_1}} \over {mn}}$$ respectively. Here m and n are constants. The relations for distance and time in two systems respectively are :

A ball is spun with angular acceleration $$\alpha$$ = 6t² $$-$$ 2t where t is in second and $$\alpha$$ is in rads^$$-$$2. At t = 0, the ball has angular velocity of 10 rads^$$-$$1 and angular position of 4 rad. The most appropriate expression for the angular position of the ball is :

A block of mass 2 kg moving on a horizontal surface with speed of 4 ms^$$-$$1 enters a rough surface ranging from x = 0.5 m to x = 1.5 m. The retarding force in this range of rough surface is related to distance by F = $$-$$kx where k = 12 Nm^$$-$$1. The speed of the block as it just crosses the rough surface will be :

A $$\sqrt {34} $$ m long ladder weighing 10 kg leans on a frictionless wall. Its feet rest on the floor 3 m away from the wall as shown in the figure. If E_f and F_w are the reaction forces of the floor and the wall, then ratio of $${F_w}/{F_f}$$ will be :

(Use g = 10 m/s².)