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 $$\overrightarrow{F} = (2\hat{i}+3\hat{j})N$$, it moves to a new point Q having coordinates (6,10)m in 4 sec . The average power and instanteous power at the end of 4 sec are in the ratio of :

Let the displacement from point P to point Q be $$\vec d$$. Then

$$\vec d = (6-3)\,\hat i + (10-4)\,\hat j = 3\,\hat i + 6\,\hat j\quad -(1)$$

The work done by the force $$\overrightarrow{F}=(2\hat i+3\hat j)\,$$N is given by

$$W = \overrightarrow{F}\cdot \vec d = 2\times3 + 3\times6 = 6 + 18 = 24\text{ J}\quad -(2)$$

Average power over the time interval $$\Delta t=4\,$$s is defined by

$$P_{\rm avg} = \frac{W}{\Delta t} = \frac{24}{4} = 6\text{ W}\quad -(3)$$

Since the force is constant, acceleration is

$$\vec a = \frac{\overrightarrow{F}}{m} = \frac{1}{4}(2\hat i +3\hat j) = 0.5\,\hat i + 0.75\,\hat j\;\mathrm{m/s^2}\quad -(4)$$

Assuming the body starts from rest, its velocity at $$t=4\,$$s is

$$\vec v = \vec a\,t = (0.5\times4)\,\hat i + (0.75\times4)\,\hat j = 2\,\hat i + 3\,\hat j\;\mathrm{m/s}\quad -(5)$$

Instantaneous power at $$t=4\,$$s is

$$P_{\rm inst} = \overrightarrow{F}\cdot \vec v = 2\times2 + 3\times3 = 4 + 9 = 13\text{ W}\quad -(6)$$

Hence the ratio of average power to instantaneous power is

$$P_{\rm avg} : P_{\rm inst} = 6 : 13$$

Final Answer: Option D; 6 : 13.