A) 3
B) 6
C) 9
D) Some other number
Correct Answer: C
Solution :
\[{{v}^{2}}-{{u}^{2}}=2as\] \[u=\sqrt{2as}\] \[ua\sqrt{S}\] \[{{u}^{2}}\alpha s\] \[\frac{v_{1}^{2}}{v_{2}^{2}}=\frac{{{s}_{1}}}{{{s}_{2}}}\] \[\frac{{{v}^{2}}}{{{v}^{2}}}=\frac{{{s}_{1}}}{{{s}_{2}}}\] \[\frac{1}{9}=\frac{{{s}_{1}}}{{{s}_{2}}}\] \[{{s}_{2}}=9{{s}_{1}}\] so the distance needed for stopping it by 9.
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