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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