1. Given data

• Mass of body: $m=2\,\text{kg}$
• Force: $\vec F=(6t\,\hat i+6t^2\,\hat j)\,\text{N}$
• The body begins to move, so initial velocity is zero: $\vec v(0)=0$

We need the instantaneous power: $P=\vec F\cdot \vec v$

2. Find acceleration using Newton's second law

$\vec a=\frac{\vec F}{m}$

So, $\vec a=\frac{1}{2}(6t\,\hat i+6t^2\,\hat j)$ $\vec a=(3t\,\hat i+3t^2\,\hat j)$

Thus,

• $a_x=3t$
• $a_y=3t^2$

3. Integrate acceleration to get velocity

Since $\vec v(0)=0$,

For the $x$-component: $v_x=\int 3t\,dt=\frac{3}{2}t^2$

For the $y$-component: $v_y=\int 3t^2\,dt=t^3$

Hence, $\vec v=\left(\frac{3}{2}t^2\,\hat i+t^3\,\hat j\right)$

4. Compute power

Instantaneous power is $P=\vec F\cdot \vec v$

Substitute: $P=(6t\,\hat i+6t^2\,\hat j)\cdot \left(\frac{3}{2}t^2\,\hat i+t^3\,\hat j\right)$

Now dot product: $P=6t\left(\frac{3}{2}t^2\right)+6t^2(t^3)$ $P=9t^3+6t^5$

So the required power is $\boxed{(9t^3+6t^5)\,\text{W}}$

5. Match with options

• A: $(3t^3+6t^5)$ ❌
• B: $(9t^5+6t^3)$ ❌
• C: $(6t^4+9t^5)$ ❌
• D: $(9t^3+6t^5)$ ✅

Therefore, the correct option is: $\boxed{\text{D}}$

6. Comparison with stored correct answer

Stored correct answer: $\text{D}$

Our derived answer is also $\text{D}$, so they agree.