A body of mass 4 kg is placed on a plane at a point $P$ having coo-dinate $(3,4) \mathrm{m}$. Under

Question

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

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(b) : Displacement 
$ P Q=(6 \hat{i}+10 \hat{j})-(3 \hat{i}+4 \hat{j})=(3 \hat{i}+6 \hat{j}) $
 Average power 
$ \begin{aligned} & \langle Pangle=\frac{W}{t}=\frac{\vec{F} \cdot \vec{S}}{t}=\frac{(2 \hat{i}+3 \hat{j})(3 \hat{i}+6 \hat{j})}{4} \ & \langle Pangle=6 \mathrm{~W} \ & \vec{a}=\frac{\vec{F}}{m}=\left(\frac{1}{2} \hat{i}+\frac{3}{4} \hat{j}ight) \mathrm{m} / \mathrm{s}^{2} \end{aligned} $
 Velocity at the end of 4 sec ,\ $\vec{v}=\left(\frac{1}{2} \hat{i}+\frac{3}{4} \hat{j}ight) 	imes 4=2 \hat{i}+3 \hat{j}$\ $P_{	ext {ins }}=\vec{F} \cdot \vec{v}=(2 \hat{i}+3 \hat{j})(2 \hat{i}+3 \hat{j})=13$\ or $\frac{\langle Pangle}{P_{	ext {ins }}}=\frac{6}{13}$\