A) 1:1
B) 1: 4
C) 1:8
D) 1:16
Correct Answer: B
Solution :
For first car, \[{{u}_{1}}=u,\] \[{{v}_{1}}=0,\] \[{{t}_{1}}=t\] \[\therefore \] \[{{v}_{1}}={{u}_{1}}+{{a}_{1}}t\] \[0=u-{{a}_{1}}t\] \[\therefore \] \[u={{a}_{1}}t\] Now, \[v_{1}^{2}=u_{1}^{2}+2{{a}_{1}}{{s}_{1}}\] \[0={{u}^{2}}-2{{a}_{1}}{{s}_{1}}\] \[{{u}^{2}}=2{{a}_{1}}{{s}_{1}}\] \[\Rightarrow \] \[{{u}^{2}}=2\times \frac{u}{t}\times {{s}_{1}}\] \[\therefore \] \[{{s}_{1}}=\frac{ut}{2}\] ??(ii) For second car, \[{{u}_{2}}=4u,\] \[{{v}_{2}}=0\] \[\therefore \] \[{{v}_{2}}={{u}_{2}}+at\] \[0=4u-{{a}_{2}}t\] \[{{a}_{2}}=\frac{4u}{t}\] ??(iii) Now, \[v_{2}^{2}=u_{2}^{2}+2{{a}_{2}}{{s}_{2}}\] \[0=u_{2}^{2}-2{{a}_{2}}{{s}_{2}}\] \[u_{2}^{2}=2{{a}_{2}}{{s}_{2}}\] \[{{(4u)}^{2}}=2\times \frac{4u}{t}\times {{s}_{2}}\] \[\therefore \] \[{{s}_{2}}=\frac{4ut}{2}\] \[\therefore \] \[\frac{{{s}_{1}}}{{{s}_{2}}}=\frac{ut/2}{4ut/2}=\frac{1}{4}\]
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