A) 12 m
B) 18 m
C) 24 m
D) 6 m
Correct Answer: C
Solution :
From third equation of motion \[{{v}^{2}}={{u}^{2}}+2a\,s\] Given \[U=50\,km/hr=50\times \frac{5}{18}m/s\] also \[V=0,\,\,{{S}_{1}}=6m\] so \[a={{\left( 50\times \frac{5}{18} \right)}^{2}}/2\times 6\] \[=\frac{{{\left( 50\times \frac{5}{18} \right)}^{2}}}{12}m/{{s}^{2}}\] Now \[U=100\,km/hr=100\times \frac{5}{18}m/\sec ;\,\,v=0\,{{S}_{2}}=?\] \[{{S}_{2}}=\frac{{{\left( 100\times \frac{5}{18} \right)}^{2}}}{2\times {{\left( 100\times \frac{5}{18} \right)}^{2}}/12}\] \[=\frac{{{(100)}^{2}}\times {{\left( \frac{5}{18} \right)}^{2}}}{2\times {{(50)}^{2}}{{\left( \frac{5}{18} \right)}^{2}}/12}=\frac{10000}{2\times 2500/12}\] \[=\frac{600}{25}=24m\]
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