A) \[({{v}_{1}}-{{v}_{2}}):({{v}_{2}}-{{v}_{3}})=({{t}_{1}}-{{t}_{2}}):({{t}_{2}}+{{t}_{3}})\]
B) \[({{v}_{1}}-{{v}_{2}}):({{v}_{2}}-{{v}_{3}})=({{t}_{1}}+{{t}_{2}}):({{t}_{2}}+{{t}_{3}})\]
C) \[({{v}_{1}}-{{v}_{2}}):({{v}_{2}}-{{v}_{3}})=({{t}_{1}}-{{t}_{2}}):({{t}_{1}}-{{t}_{3}})\]
D) \[({{v}_{1}}-{{v}_{2}}):({{v}_{2}}-{{v}_{3}})=({{t}_{1}}-{{t}_{2}}):({{t}_{2}}-{{t}_{3}})\]
Correct Answer: B
Solution :
Let \[{{u}_{1}},\,{{u}_{2}},\,{{u}_{3}}\] and \[{{u}_{4}}\] be velocities at time \[t=0,\ {{t}_{1}},\ ({{t}_{1}}+{{t}_{2}})\] and \[({{t}_{1}}+{{t}_{2}}+{{t}_{3}})\]respectively and acceleration is a then \[{{v}_{1}}=\frac{{{u}_{1}}+{{u}_{2}}}{2},\ {{v}_{2}}=\frac{{{u}_{2}}+{{u}_{3}}}{2}\text{and}\ {{v}_{3}}=\frac{{{u}_{3}}+{{u}_{4}}}{2}\] Also \[{{u}_{2}}={{u}_{1}}+a{{t}_{1}}\ ,\ {{u}_{3}}={{u}_{1}}+a({{t}_{1}}+{{t}_{2}})\] and \[{{u}_{4}}={{u}_{1}}+a({{t}_{1}}+{{t}_{2}}+{{t}_{3}})\] By solving, we get \[\frac{{{v}_{1}}-{{v}_{2}}}{{{v}_{2}}-{{v}_{3}}}=\frac{({{t}_{1}}+{{t}_{2}})}{({{t}_{2}}+{{t}_{3}})}\]
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