A) \[1:1\]
B) \[1:4\]
C) \[1:8\]
D) \[1:16\]
Correct Answer: D
Solution :
| [d] In this question, the cars are identical means coefficient of friction between the tyre and the ground is same for both the cars, as a result retardation is same for both the cars equal to\[\mu g\]. \[(\because mg=\mu mg\Rightarrow a=\mu g)\] |
| Let first car travel distance \[{{s}_{1}}\], before stopping while second car travel distance \[{{s}_{2}}\], then from |
| \[{{v}^{2}}={{u}^{2}}-2\] as, we get |
| \[O={{u}^{2}}-2\mu g\times {{s}_{1}}\] |
| \[\Rightarrow \] \[{{s}_{1}}=\frac{{{u}^{2}}}{2\mu g}\] |
| and \[0={{(4u)}^{2}}-2\mu g\times {{s}_{2}}\] |
| \[\Rightarrow \] \[{{s}_{2}}=\frac{16{{u}^{2}}}{2\mu g}=16{{s}_{1}}\] |
| \[\Rightarrow \] \[\frac{{{s}_{1}}}{{{s}_{2}}}=\frac{1}{16}\] |
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