• # question_answer                 A car moving with a speed of 40 km/h can be stopped by applying brakes after at least 2 m. If me same car is moving with a speed of 80 km/h, what is the minimum stopping distance?           A)                 8 m                         B)                 2 m         C)                            4 m     D)                                            6 m

According to conservation of energy, the kinetic energy of car = work done in stopping the car                 i.e.,        $\frac{1}{2}m{{v}^{2}}=Fs$                 where F is the retarding force and s the stopping distance.                 For same retarding force, $s\propto \,\,{{v}^{2}}$                 $\therefore \frac{{{s}_{2}}}{{{s}_{1}}}={{\left( \frac{{{v}_{2}}}{{{v}_{1}}} \right)}^{2}}={{\left( \frac{80}{40} \right)}^{2}}=4$                 $\therefore {{s}_{2}}=4{{s}_{1}}=4\times 2=8\,m$                 Alternative: Initial speed of cat                 $u=40\,km/h=40\times \frac{5}{18}m/s=\frac{100}{9}m/s$                 From 3rd equation of motion                 ${{v}^{2}}={{u}^{2}}-2as$ $\Rightarrow$               $0={{\left( \frac{100}{9} \right)}^{2}}-2\times a\times 2$ $\Rightarrow$               $a=\frac{2500}{81}m/{{s}^{2}}$                 Final speed of car = 80 km/h                 $=80\times \frac{5}{18}=\frac{200}{9}m/s$                 Suppose car stops for a distance $s'$. Then                 ${{v}^{2}}={{u}^{2}}-2as'$                 $0={{\left( \frac{200}{9} \right)}^{2}}-2\times \frac{2520}{81}s'$                 $\Rightarrow$               $s'=\frac{200\times 200\times 81}{9\times 9\times 2\times 2500}=8\,m$