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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