A) \[({{v}_{1}}-{{v}_{2}}):({{v}_{2}}-{{v}_{3}})=({{t}_{1}}-{{t}_{2}}):({{t}_{1}}-{{t}_{3}})\]
B) \[({{v}_{1}}-{{v}_{2}}):({{v}_{2}}-{{v}_{3}})=({{t}_{1}}-{{t}_{2}}):({{t}_{2}}-{{t}_{3}})\]
C) \[({{v}_{1}}-{{v}_{2}}):({{v}_{2}}-{{v}_{3}})=({{t}_{1}}+{{t}_{2}}):({{t}_{2}}+{{t}_{3}})\]
D) None of the above
Correct Answer: C
Solution :
Let O be starting point and AB and C are three successive positions suppose velocity at \[0 = zero\] as average velocity in internal \[{{t}_{1}}\,\,is\,\,{{v}_{1}}\] \[\therefore \,Velocity at\,\,A =\,\,{{v}_{1}}\] as average velocity in internal \[{{t}_{2}}\,\,is\,\,{{v}_{2}}\] as average velocity in internal \[{{t}_{3}}\,\,is\,\,{{v}_{3}}\] \[\therefore \,\,\,Velocity at C = \left[ {{v}_{3}}-\left( {{v}_{2}}-{{v}_{1}} \right) \right]\] \[=\,\,\,\,{{v}_{3}}-{{v}_{2}}+{{v}_{1}}\] Now, using \[\operatorname{v} =u + at\] we have \[{{v}_{1}}=0+a{{t}_{1}}\] \[\Rightarrow \,\,\,\,\,\,\,\,\,\,{{v}_{1}}=a{{t}_{1}}\] ? (i) Similarly, \[({{v}_{2}}-{{v}_{1}})=a({{t}_{1}}+{{t}_{2}})\] ? (ii) \[({{v}_{3}}-{{v}_{2}}+{{v}_{1}})\,\,=\,\,a({{t}_{1}}+{{t}_{2}}+{{t}_{3}})\] ? (iii) From Eq. (i) and Eq. (iii), we get \[{{v}_{3}}-{{v}_{2}}=a({{t}_{2}}+{{t}_{3}})\] ? (iv) Dividing Eq. (ii) by Eq. (iv) \[\frac{({{v}_{2}}-{{v}_{1}})}{({{v}_{3}}-{{v}_{2}})}=\frac{a}{a}\,\frac{({{t}_{1}}+{{t}_{2}})}{({{t}_{2}}+{{t}_{3}})}\] \[\frac{{{v}_{1}}-{{v}_{2}}}{{{v}_{2}}-{{v}_{3}}}\,\,=\,\,\frac{{{t}_{1}}+{{t}_{2}}}{{{t}_{2}}+{{t}_{3}}}\]
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