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 \[{{\text{v}}_{\text{av}}}\text{=}\frac{\text{total}\,\text{distance covered}}{\text{total time elapased}}\] \[=\frac{{{v}_{1}}{{t}_{1}}+{{v}_{2}}{{t}_{2}}+{{v}_{3}}{{t}_{3}}+...}{{{t}_{1}}+{{t}_{2}}+{{t}_{3}}+...}\] 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 \[{{v}_{av}}=\frac{{{v}_{1}}+{{v}_{2}}}{2}\]
You need to login to perform this action.
You will be redirected in
3 sec
