An automobile travelling with a speed of $$60 \,km/h,$$  can brake to stop within a distance of $$20 \,m.$$  If the car is going twice as fast i.e., $$120 \,km/h,$$  the stopping distance will be-

Solution :
$$\eqalign{ & {\text{Speed,}}\,\,u = 60 \times \frac{5}{{18}}\,m/s = \frac{{50}}{3}\,m/s \cr & d = 20\,m,\,\,u' = 120 \times \frac{5}{{18}} = \frac{{100}}{3}\,m/s \cr & {\text{Let declearation be a then }}{\left( 0 \right)^2} - {u^2} = - 2ad \cr & {\text{or, }}{\left( 0 \right)^2} - {u^2} = - 2ad\,\,\,\,{\text{or}},\,\,{u^2} = 2ad\,.....(1) \cr & {\text{and }}{\left( 0 \right)^2} - u{'^2} = - 2ad'\,\,\,\,\,{\text{or}},\,\,u{'^2} = 2ad'.....\left( 2 \right) \cr & (2)\;{\text{divided }}(1)\,{\text{gives,}} \cr & 4 = \frac{{d'}}{d} \Rightarrow d' = 4 \times 20 = 80\,m \cr} $$