J & K CET Engineering J and K - CET Engineering Solved Paper-2014

  • question_answer
    The bus moving with a speed of \[42\text{ }km/h\]is brought to a stop by brakes after\[6\text{ }m\]. If the same bus is moving at a speed of \[90\text{ }km/h,\] then the minimum stopping distance is

    A)  \[15.48\text{ }m\]

    B)  \[18.64\text{ }m\]

    C)  \[22.13\text{ }m\]

    D)  \[27.55\text{ }m\]

    Correct Answer: D

    Solution :

    \[u=42\times \frac{5}{18}=11.66m/s\] and  \[v=0\] So,     \[{{s}_{1}}=6m\] when,  \[u=90\times \frac{5}{18}=25m/s\] and     \[v=0\]        . Then, \[{{s}_{2}}=?\] For first case \[\Rightarrow \] \[{{v}^{2}}={{u}^{2}}+2a{{s}_{1}}\] \[0={{(11.66)}^{2}}+2a\times 6\] \[({{s}_{1}}=6m)\] \[a=\frac{-11.66\times 11.66}{12}\] \[a=-11.33m/{{s}^{2}}\] For second case \[{{v}^{2}}={{u}^{2}}+2a{{s}_{2}}\] \[0={{(25)}^{2}}+2(-11.33)\times {{s}_{2}}\] \[{{s}_{2}}=\frac{625}{2\times 11.33}=27.5m\]


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