A body of mass 4 kg is placed on a plane at a point P having coordinate (3, 4) m. Under the action of force $\vec{F} = (2 \hat{i} \to + 3 \hat{j}) N,$ it moves to a new point Q having coordinates (6, 10)m in 4 sec. The average power and instantaneous power at the end of 4 sec are in the ratio of :

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4 : 3

1 : 2

13 : 6

6 : 13

answer is B.

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Detailed Solution

$\begin{array}{l} < p > = \frac{(2 \hat{i} + 3 \hat{j}) \cdot (3 \hat{i} + 6 \hat{j})}{4} = 6 \\ \vec{a} = (\frac{\vec{F}}{m} = \frac{1}{2} \hat{i} + \frac{3}{4} \hat{j}) \end{array}$
$\begin{array}{l} \vec{v} at t = 4 \sec = (\frac{1}{2} \hat{i} + \frac{3}{4} \hat{j}) \times 4 = (2 \hat{i} + 3 \hat{j}) \\ P_{ins} = (2 \hat{i} + 3) (2 \hat{i} + 3 \hat{j}) = 13 \\ \frac{< P >}{P_{ins}} = \frac{6}{13} \end{array}$
Note : Given data is not matching.
$\begin{array}{l} S = ut + \frac{1}{2} {at}^{2} \\ S = 0 + \frac{1}{2} \frac{(2 \hat{i} + 3 \hat{j})}{4} (4)^{2} = 4 \hat{i} + 6 \hat{j} \end{array}$
If ${\vec{r}}_{i} = 3 \hat{i} + 4 \hat{j} then {\vec{r}}_{f} = 7 \hat{i} + 10 \hat{j}$
But Final position given in the question is (6, 10).