JEE Main & Advanced Physics Vectors Question Bank Fundamentals of Vectors

  • question_answer The position vector of a particle is determined by the expression \[\vec{r}=3{{t}^{2}}\hat{i}+4{{t}^{2}}\hat{j}+7\hat{k}\] The distance traversed in first 10 sec is                [DPMT 2002]

    A)            500 m                                      

    B)            300 m

    C)            150 m                                      

    D)            100 m

    Correct Answer: A

    Solution :

                 \[\vec{r}=3{{t}^{2}}\hat{i}+4{{t}^{2}}\hat{j}+7\hat{k}\]                    at \[t=0\], \[{{\vec{r}}_{1}}=7\hat{k}\]                    at \[t=10\sec \], \[{{\vec{r}}_{2}}=300\hat{i}+400\hat{j}+7\hat{k}\],                    \[\overrightarrow{\Delta r}={{\vec{r}}_{2}}-{{\vec{r}}_{1}}=300\hat{i}+400\hat{j}\]                    \[|\overrightarrow{\Delta r}|\,=\,|{{\vec{r}}_{2}}-{{\vec{r}}_{1}}|\,=\sqrt{{{(300)}^{2}}+{{(400)}^{2}}}=500m\]


You need to login to perform this action.
You will be redirected in 3 sec spinner