STOPPING distance, \[S\propto {{U}^{2}}\] \[(\because \ {{v}^{2}}={{u}^{2}}-2AS)\]             Þ \[\frac{{{S}_{2}}}{{{S}_{1}}}={{\LEFT( \frac{{{u}_{2}}}{{{u}_{1}}} \right)}^{2}}={{\left( \frac{120}{60} \right)}^{2}}=4\]             Þ \[{{S}_{2}}=4\times {{S}_{1}}=4\times 20=80\,m\]

"> STOPPING distance, \[S\propto {{U}^{2}}\] \[(\because \ {{v}^{2}}={{u}^{2}}-2AS)\]             Þ \[\frac{{{S}_{2}}}{{{S}_{1}}}={{\LEFT( \frac{{{u}_{2}}}{{{u}_{1}}} \right)}^{2}}={{\left( \frac{120}{60} \right)}^{2}}=4\]             Þ \[{{S}_{2}}=4\times {{S}_{1}}=4\times 20=80\,m\]

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An automobile travelling with a speed of \[60\,\,km/h,\] can brake to stop within a distance of 20 m. If the car is going twice as fast, i.e. 120 km/h, the stopping distance will be [AIEEE 2004]

Joint Entrance Exam - Main (JEE Main) Physics in Joint Entrance Exam - Main (JEE Main) . 10 months ago

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The STOPPING distance, \[S\propto {{U}^{2}}\] \[(\because \ {{v}^{2}}={{u}^{2}}-2AS)\]             Þ \[\frac{{{S}_{2}}}{{{S}_{1}}}={{\LEFT( \frac{{{u}_{2}}}{{{u}_{1}}} \right)}^{2}}={{\left( \frac{120}{60} \right)}^{2}}=4\]             Þ \[{{S}_{2}}=4\times {{S}_{1}}=4\times 20=80\,m\]

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