A force $$\vec{F} = \left(40\hat{i} + 10\hat{j}\right)$$ N acts on a body of mass 5 kg. If the body starts from rest, its position vector $$\vec{r}$$ at time $$t = 10$$ s will be

We have a constant force $$\vec F = 40\hat i + 10\hat j \;{\rm N}$$ acting on a body whose mass is $$m = 5\;{\rm kg}$$.

According to Newton’s second law, the relation between force and acceleration is first stated as $$\vec F = m\,\vec a.$$

We now solve for the acceleration $$\vec a$$ by dividing the force vector by the mass:

$$\vec a = \frac{\vec F}{m} = \frac{40\hat i + 10\hat j}{5} = \left(\frac{40}{5}\right)\hat i + \left(\frac{10}{5}\right)\hat j = 8\hat i + 2\hat j\;{\rm m\,s^{-2}}.$$

The body starts from rest, so its initial velocity is

$$\vec v_0 = 0\hat i + 0\hat j\;{\rm m\,s^{-1}}.$$

For motion with constant acceleration, the position vector after time $$t$$ is given by the kinematic formula (which we state explicitly):

$$\vec r = \vec r_0 + \vec v_0\,t + \frac12\,\vec a\,t^{2}.$$

The body is released from the origin, so $$\vec r_0 = 0\hat i + 0\hat j$$. Because $$\vec v_0 = 0$$, the second term also vanishes. Therefore, the position vector simplifies to

$$\vec r = \frac12\,\vec a\,t^{2}.$$

Substituting the acceleration $$\vec a = 8\hat i + 2\hat j$$ and the time $$t = 10\;{\rm s}$$, we obtain

$$\vec r = \frac12\,(8\hat i + 2\hat j)\,(10)^{2} = \frac12\,(8\hat i + 2\hat j)\,(100).$$

Carrying out the multiplication, we get component-wise values:

$$\vec r = \left[\frac12 \times 8 \times 100\right]\hat i + \left[\frac12 \times 2 \times 100\right]\hat j = (4 \times 100)\hat i + (1 \times 100)\hat j = 400\hat i + 100\hat j\;{\rm m}.$$

Thus the particle’s position vector after 10 seconds is $$\left(400\hat i + 100\hat j\right)\,{\rm m}$$.

Hence, the correct answer is Option C.