Let's define a coordinate system where the initial position is the origin, east is the positive x-direction, and south is the positive y-direction. The orthogonal forces allow us to analyze the motion in the x and y directions independently.

[Q1] Find the magnitude of the displacement of the object during these 4 seconds.

Motion in the x-direction (East): During the first phase ($0 \le t \le t_1$), the object undergoes constant acceleration under force $F_1$. The acceleration is $a_x = F_1/m$. The displacement during this phase is:

$$d_{1x} = \frac{1}{2}a_x t_1^2 = \frac{F_1 t_1^2}{2m}$$

At time $t_1$, the velocity reaches $v_x = a_x t_1 = F_1 t_1 / m$. During the second phase ($t_1 < t \le t_1+t_2$), there is no force in the x-direction, so the object moves at a constant velocity $v_x$. The displacement during this phase is:

$$d_{2x} = v_x t_2 = \frac{F_1 t_1 t_2}{m}$$

The total x-displacement is the sum of the displacements from both phases:

$$d_x = d_{1x} + d_{2x} = \frac{F_1 t_1^2}{2m} + \frac{F_1 t_1 t_2}{m} = \frac{F_1 t_1}{m}\left(\frac{t_1}{2} + t_2\right)$$

Motion in the y-direction (South): The object is at rest in the y-direction during the first phase ($d_{1y}=0$). During the second phase ($t_1 < t \le t_1+t_2$), it starts from rest in this direction and experiences a constant acceleration under force $F_2$. The acceleration is $a_y = F_2/m$. The y-displacement during this phase is:

$$d_y = \frac{1}{2}a_y t_2^2 = \frac{F_2 t_2^2}{2m}$$

Total Displacement: The total displacement vector is $\vec{d} = d_x \hat{i} + d_y \hat{j}$. Its magnitude $d$ is found using the Pythagorean theorem:

$$d = \sqrt{d_x^2 + d_y^2}$$

Substituting the numerical values:

$m = 2$ kg, $F_1 = 6$ N, $t_1 = 2$ s, $F_2 = 8$ N, $t_2 = 2$ s.

First, calculate the component displacements:

$$d_x = \frac{6 \text{ N} \cdot 2 \text{ s}}{2 \text{ kg}}\left(\frac{2 \text{ s}}{2} + 2 \text{ s}\right) = 6(1+2) \text{ m} = 18 \text{ m}$$ $$d_y = \frac{8 \text{ N} \cdot (2 \text{ s})^2}{2 \cdot 2 \text{ kg}} = \frac{32}{4} \text{ m} = 8 \text{ m}$$

Now, calculate the magnitude of the total displacement:

$$d = \sqrt{(18 \text{ m})^2 + (8 \text{ m})^2} = \sqrt{324 + 64} \text{ m} = \sqrt{388} \text{ m}$$ $$d = \sqrt{4 \cdot 97} \text{ m} = 2\sqrt{97} \text{ m}$$