A) \[\sqrt{La}\]
B) \[\sqrt{2La}\]
C) \[\sqrt{3La}\]
D) \[\sqrt{5La}\]
Correct Answer: A
Solution :
Let v be the maximum velocity attained and t the total time of journey. t' is the duration of acceleration and retardation. Then \[v=0+at.\] \[\therefore \text{L=}\frac{\text{1}}{\text{2}}\text{at}{{\text{ }\!\!'\!\!\text{ }}^{\text{2}}}+v\left( \text{t}-2\text{t}' \right)\text{+}\frac{\text{1}}{\text{2}}\text{at}{{\text{ }\!\!'\!\!\text{ }}^{\text{2}}}\] \[\text{=a}{{\left( \frac{\text{v}}{\text{a}} \right)}^{\text{2}}}\text{+v}\left( \text{t}-\frac{\text{2v}}{\text{a}} \right)\] \[=\frac{{{\text{v}}^{\text{2}}}}{\text{a}}\text{+vt}-\frac{\text{2}{{\text{v}}^{\text{2}}}}{\text{a}}\text{=vt}-\frac{{{\text{v}}^{\text{2}}}}{\text{a}}\] \[\text{=t=}\frac{\text{L}}{\text{v}}\text{+}\frac{\text{v}}{\text{a}}\Rightarrow \frac{\text{dv}}{\text{dt}}=0\text{ }\therefore {{\text{v}}_{\max }}=\sqrt{\text{La}}\]You need to login to perform this action.
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