A) \[(\sqrt{{{a}_{1}}/{{a}_{2}}})t\]
B) \[(\sqrt{{{a}_{2}}/{{a}_{1}}})t\]
C) \[({{a}_{1}}\sqrt{{{a}^{2}}})t\]
D) \[(\sqrt{{{a}_{1}}{{a}_{2}}})t\]
Correct Answer: D
Solution :
Consider that A takes \[{{t}_{1}}\] second, then according to the given problem, B will take \[({{t}_{1}}+t)\] seconds. Further let \[{{v}_{1}}\] be the velocity of B at finishing point, then velocity of A will be. \[({{v}_{1}}+v)\]Writing equations of motion for A and B \[{{v}_{1}}+v={{a}_{1}}{{t}_{1}}\] ? (i) \[{{v}_{1}}={{a}_{2}}({{t}_{1}}+t)\] ? (ii) From Eqs.(i) and (ii),we get \[v=({{a}_{1}}-{{a}_{2}}){{t}_{1}}-{{a}_{2}}t\] ? (iii) Total distance travelled by both the cars is equal \[{{s}_{A}}={{s}_{B}}\] \[\Rightarrow \] \[\frac{1}{2}{{a}_{1}}t_{1}^{2}=\frac{1}{2}{{a}_{2}}{{({{t}_{1}}+t)}^{2}}\Rightarrow {{t}_{1}}=\frac{{{\sqrt{a}}_{2}}t}{\sqrt{{{a}_{1}}}-\sqrt{{{a}_{2}}}}\]Substituting this value of \[{{t}_{1}}\] in Eq. (iii), we get \[v=(\sqrt{{{a}_{1}}{{a}_{2}}})t\]
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