A) \[\frac{S}{v}+\frac{v}{2}\left( \frac{1}{\alpha }+\frac{1}{\beta } \right)\]
B) \[\frac{S}{v}+\frac{v}{\alpha }+\frac{v}{\beta }\]
C) \[\left( \frac{v}{\alpha }+\frac{v}{\beta } \right)\]
D) \[\frac{S}{v}-\frac{v}{2}\left( \frac{v}{\alpha }+\frac{1}{\beta } \right)\]
Correct Answer: A
Solution :
From v = u + at, we have \[\text{v}=0+\alpha {{\text{t}}_{1}}\Rightarrow \frac{\text{v}}{\alpha }\] \[0=v-\beta {{t}_{3}}\Rightarrow {{t}_{3}}=\frac{\text{v}}{\beta }.\] \[\text{Now, S =}\frac{\text{1}}{\text{2}}\text{v}{{\text{t}}_{\text{1}}}\text{+v}{{\text{t}}_{\text{2}}}+\frac{1}{2}\text{v}{{\text{t}}_{\text{3}}}=\frac{{{\text{v}}^{\text{2}}}}{2\alpha }+\text{v}{{\text{t}}_{2}}+\frac{{{\text{v}}^{\text{2}}}}{2\beta }\] \[\Rightarrow {{\text{t}}_{\text{2}}}\text{=}\frac{\text{S}}{\text{v}}-\frac{\text{v}}{\text{2}}\left( \frac{1}{\alpha }+\frac{1}{\beta } \right).\]\[\text{Hence, }{{\text{t}}_{\text{1}}}\text{+}{{\text{t}}_{\text{2}}}\text{+}{{\text{t}}_{\text{3}}}\text{=}\frac{\text{S}}{\text{v}}\text{+}\frac{\text{v}}{\text{2}}\left( \frac{1}{\alpha }+\frac{1}{\beta } \right)\]You need to login to perform this action.
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