A) 20m
B) 40m
C) 60 m
D) 80 m
Correct Answer: D
Solution :
The braking retardation will remain same and assumed to be constant, let it be\[a\]. From 3rd equation of motion,\[{{v}^{2}}={{u}^{2}}+2as\] 1st case: \[0={{\left( 60\times \frac{5}{18} \right)}^{2}}-2a\times {{s}_{1}}\] \[\Rightarrow \] \[{{s}_{1}}=\frac{{{(60\times 5/18)}^{2}}}{2a}\] 2nd case: \[0={{\left( 120\times \frac{5}{18} \right)}^{2}}-2a\times {{s}_{2}}\] \[\Rightarrow \] \[{{s}_{2}}=\frac{{{\left( 120\times 5/18 \right)}^{2}}}{2a}\] \[\therefore \]\[\frac{{{s}_{1}}}{{{s}_{2}}}=\frac{1}{4}\Rightarrow {{s}_{2}}=4{{s}_{1}}=4\times 20=80\,m\]
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