• 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

    Correct Answer: A

    Solution :

    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                 \[{{v}^{2}}={{u}^{2}}-2as'\]                 \[0={{\left( \frac{200}{9} \right)}^{2}}-2\times \frac{2520}{81}s'\]                 \[\Rightarrow \]               \[s'=\frac{200\times 200\times 81}{9\times 9\times 2\times 2500}=8\,m\]

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