A) \[\sqrt{{{v}_{1}}{{v}_{2}}}\]
B) \[\left( \frac{{{v}_{1}}+{{v}_{2}}}{2} \right)\]
C) \[{{\left( \frac{1}{{{v}_{1}}}+\frac{1}{{{v}_{2}}} \right)}^{-1}}\]
D) \[2{{\left( \frac{1}{{{v}_{1}}}+\frac{1}{{{v}_{2}}} \right)}^{-1}}\]
Correct Answer: B
Solution :
When particle moves with different uniform speed \[{{v}_{1}},{{v}_{2}}{{v}_{3}}....\] etc in different time intervals\[{{t}_{1}},{{t}_{2}}{{t}_{3}},....\].. etc respectively, its average speed over the total time of journey is given as \[{{v}_{av}}=\frac{total\text{ }distance\text{ }covered}{total\text{ }time\text{ }elapsed}\] \[\frac{{{v}_{1}}{{t}_{1}}+{{v}_{2}}{{t}_{2}}+{{v}_{3}}{{t}_{3}}+....}{{{t}_{1}}+{{t}_{2}}+{{t}_{3}}+....}\] Here, \[{{v}_{av}}=\frac{{{v}_{1}}{{t}_{1}}+{{v}_{2}}{{t}_{2}}}{{{t}_{1}}+{{t}_{2}}}\] \[=\frac{({{v}_{1}}+{{v}_{2}})t}{2t}\] \[[\because {{t}_{1}}={{t}_{2}}]\] or \[{{v}_{av}}=\frac{{{v}_{1}}+{{v}_{2}}}{2}\]
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