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AMU Medical AMU Solved Paper-2012

  • question_answer
    Let \[{{r}_{1}}\,(t)=3\,t\,i+4{{t}^{2}}j\] and \[{{r}_{2}}\,(t)=4\,{{t}^{2}}\,i+3r\,\,j\] represent the positions of particles 1 and 2, respectively, as function of time \[t,\,\,{{r}_{1}}(t)\] and \[{{r}_{2}}(t)\] are in metre and t in second. The relative speed of the two particles at the instant t = 1 s, will be

    A)  1 m/s                                   

    B)  \[3\sqrt{2}\,m/s\]

    C)  \[5\sqrt{2}\,m/s\]         

    D)  \[7\sqrt{2}\,m/s\]

    Correct Answer: C

    Solution :

                     Given,                 \[{{r}_{1}}(t)=3\,t\,i+4\,{{t}^{2}}j\] \[\therefore \]  \[\frac{d{{r}_{1}}}{dt}=3i+8\,tj\] At           \[t=1s\]                 \[{{v}_{1}}=\frac{d{{r}_{1}}}{dt}=3i+8j\] Again, \[{{r}_{2}}(t)=4\,{{t}^{2}}i+3tj\]                 \[\frac{d{{r}_{2}}}{dt}=8ti+3j\] At           \[t=1s\]                 \[{{v}_{2}}=\frac{d{{r}_{2}}}{dt}=8i+3j\] Relative velocity \[={{v}_{1}}-{{v}_{2}}\]                                                 \[=-5i=5j\]                                                 \[=\sqrt{{{(5)}^{2}}+{{(5)}^{2}}}=5\sqrt{2}\,m/s\]


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