An automobile moves on a road with a speed of $$54\,km{h^{ - 1}}.$$  The radius of its wheels is $$0.45\,m$$  and the moment of inertia of the wheel about its axis of rotation is $$3\,kg\,{m^2}.$$  If the vehicle is brought to rest in $$15\,s,$$  the magnitude of average torque transmitted by its brakes to the wheel is

Solution :
As velocity of an automobile vehicle,
$$v = 54\,km/h = 54 \times \frac{5}{{18}} = 15\,m/s$$
Angular velocity of a vehicle, $$v = {\omega _0}r$$
$$ \Rightarrow {\omega _0} = \frac{V}{R} = \frac{{15}}{{0.45}} = \frac{{100}}{3}rad/s$$
So, angular acceleration of an automobile,
$$\alpha = \frac{{\Delta \omega }}{t} = \frac{{{\omega _f} - {\omega _0}}}{t} = \frac{{0 - \frac{{100}}{3}}}{{15}} = \frac{{ - 100}}{{45}}rad/{s^2}$$
Thus, average torque transmitted by its brakes to wheel
$$\tau = I\alpha \Rightarrow 3 \times \frac{{100}}{{45}} = 6.66\,kg{m^2}{s^{ - 2}}$$