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9th Class Science Time and Motion Question Bank Motion IIT JEE Objective Problems

  • question_answer
    A point moves with uniform acceleration\[{{v}_{1}},{{v}_{2}}\]and\[{{v}_{3}}\]. Denote the average velocities in three successive intervals of time\[{{t}_{1}},{{t}_{2}}\]and\[{{t}_{3}}\].Correct relation among the following is

    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)  \[({{v}_{1}}-{{v}_{2}}):({{v}_{2}}-{{v}_{3}})=({{t}_{1}}-{{t}_{2}}):({{t}_{2}}+{{t}_{3}})\]

    D)  \[({{v}_{1}}-{{v}_{2}}):({{v}_{2}}-{{v}_{3}})=({{t}_{1}}+{{t}_{2}}):({{t}_{2}}-{{t}_{3}})\]

    Correct Answer: D

    Solution :

    Let u be the initial velocity \[\therefore {{v}_{1}}'=u+a{{t}_{1}},\,\,{{v}_{2}}'=u+a({{t}_{1}}+{{t}_{2}})\]and \[{{v}_{3}}'=u+a({{t}_{1}}+{{t}_{2}}+{{t}_{3}})\] Now \[{{V}_{1}}=\frac{u+V_{1}^{'}}{2}=\frac{u+(u+a{{t}_{1}})}{2}=u+\frac{1}{2}a{{t}_{1}}\] \[{{V}_{2}}=\frac{V_{1}^{'}+V_{2}^{'}}{2}=u+a{{t}_{1}}+\frac{1}{2}a{{t}_{2}}\] \[{{V}_{3}}=\frac{V_{2}^{'}+V_{3}^{'}}{2}=u+a{{t}_{1}}+a{{t}_{2}}+\frac{1}{2}a{{t}_{3}}\] So, \[{{v}_{1}}-{{v}_{2}}=-\frac{1}{2}a({{t}_{1}}+{{t}_{2}})\]and \[{{v}_{2}}-{{v}_{3}}=-\frac{1}{2}a({{t}_{2}}+{{t}_{3}})\] \[\therefore ({{V}_{1}}-{{V}_{2}}):({{V}_{2}}-{{V}_{3}})=({{t}_{1}}+{{t}_{2}}):({{t}_{2}}+{{t}_{3}})\]


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