CEE Kerala Engineering CEE Kerala Engineering Solved Paper-2010

  • question_answer
    The position of a particle is given by \[\overrightarrow{r}=2{{t}^{2}}\hat{i}+3t\hat{j}+4\hat{k},\]where t is in second and the coefficients have proper units for\[\overrightarrow{r}\]to be in metre. The\[\overrightarrow{a}(t)\]of the particle at\[t=1\text{ }s\] is

    A)  \[4\text{ }m{{s}^{-2}}\]along \[\text{y-}\]direction

    B)  \[\text{3 }m{{s}^{-2}}\]along\[x\text{-}\]direction

    C)  \[4\text{ }m{{s}^{-2}}\]along\[x\text{-}\]direction

    D)  \[\text{2 }m{{s}^{-2}}\]along\[\text{z-}\]direction

    E) \[\text{3 }m{{s}^{-2}}\]along\[\text{z-}\]direction

    Correct Answer: C

    Solution :

    Distance, \[\overrightarrow{\mathbf{r}}=2{{t}^{2}}\mathbf{\hat{i}}+3t\mathbf{\hat{j}}+4\mathbf{\hat{k}}\] Velocity, \[\overrightarrow{\mathbf{v}}=\frac{\overrightarrow{\mathbf{dr}}}{dt}=4t\mathbf{\hat{i}}+3\mathbf{\hat{j}}\] Acceleration, \[a=\frac{dv}{dt}=4\mathbf{\hat{i}}\] \[=4\text{ }m/{{s}^{2}}\]in the\[x-\]direction.


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