A) \[({{v}_{1}}{{v}_{2}}):({{v}_{2}}{{v}_{3}})=({{t}_{1}}{{t}_{2}}):\text{(}{{t}_{2}}+{{t}_{3}})\]
B) \[({{v}_{1}}{{v}_{2}}):({{v}_{2}}{{v}_{3}})=({{t}_{1}}+{{t}_{2}}):\text{(}{{t}_{2}}+{{t}_{3}})\]
C) \[({{v}_{1}}{{v}_{2}}):({{v}_{2}}{{v}_{2}})=({{t}_{1}}-{{t}_{2}}):\text{(}{{t}_{2}}-{{t}_{3}})\]
D) \[({{v}_{1}}{{v}_{2}}):({{v}_{2}}+{{v}_{3}})=({{t}_{1}}-{{t}_{2}}):\text{(}{{t}_{2}}+{{t}_{3}})\]
Correct Answer: B
Solution :
| [b] Suppose velocity at O = zero |
| As average velocity in interval \[{{t}_{1}}\] is \[{{v}_{1}}\]. |
| velocity at \[A={{v}_{1}}\] |
| As average velocity in interval \[{{t}_{2}}\] is \[{{v}_{2}},\] |
| velocity at \[B=({{v}_{2}}{{v}_{1}})\] | |
| As average velocity in interval \[{{t}_{3}}\] is \[{{v}_{3}},\] velocity at \[C\text{ }({{v}_{3}}{{v}_{2}}+\text{ }{{v}_{1}})\] | |
| Using \[v=u+at\] | |
| \[{{v}_{1}}=0+a{{t}_{1}}\] | ....(i) |
| \[({{v}_{2}}{{v}_{1}})=0+a({{t}_{1}}+\text{ }{{t}_{2}})\] | ....(ii) |
| \[({{v}_{3}}{{v}_{2}}+{{v}_{1}})=0+a({{t}_{1}}+{{t}_{2}}+{{t}_{3}})\] | ....(iii) |
| Subtract Eq. (i) from Eq. (iii) | |
| \[({{v}_{3}}{{v}_{2}})=a({{t}_{2}}+{{t}_{3}})\] | ....(iv) |
| Divided Eq. (ii) by Eq. (iv) | |
| \[\frac{({{v}_{2}}-\,{{v}_{1}})}{({{v}_{3}}-{{v}_{2}})}\,=\,\frac{a({{t}_{1}}+{{t}_{2}})}{a({{t}_{2}}+{{t}_{3}})}\] | |
| \[\frac{({{v}_{1}}-\,{{v}_{2}})}{({{v}_{2}}-{{v}_{3}})}\,=\,\frac{{{t}_{1}}+{{t}_{2}}}{{{t}_{2}}+{{t}_{3}}}\] | |
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