The figure shows $(v_x, t)$ and $(v_y, t)$ diagrams for a body of unit mass $(m = 1 \text{ kg})$. Find the force as a function of time.

(N/A) Given: Mass $m = 1 \text{ kg}$.
Force $\vec{F} = m\vec{a} = 1 \cdot (a_x \hat{i} + a_y \hat{j}) = a_x \hat{i} + a_y \hat{j}$.
For $v_x - t$ graph $(a)$:
$1$. For $0 \le t \le 1 \text{ s}$,slope $a_x = \frac{2 - 0}{1 - 0} = 2 \text{ m/s}^2$. Thus,$F_x = 1 \times 2 = 2 \text{ N}$.
$2$. For $1 \le t \le 2 \text{ s}$,slope $a_x = \frac{0 - 2}{2 - 1} = -2 \text{ m/s}^2$. Thus,$F_x = 1 \times (-2) = -2 \text{ N}$.
For $v_y - t$ graph $(b)$:
$1$. For $0 \le t \le 1 \text{ s}$,slope $a_y = \frac{1 - 0}{1 - 0} = 1 \text{ m/s}^2$. Thus,$F_y = 1 \times 1 = 1 \text{ N}$.
$2$. For $t > 1 \text{ s}$,velocity is constant,so $a_y = 0$. Thus,$F_y = 0 \text{ N}$.
Resultant Force $\vec{F}(t)$:
- For $0 \le t \le 1 \text{ s}: \vec{F} = 2\hat{i} + 1\hat{j} \text{ N}$.
- For $1 < t \le 2 \text{ s}: \vec{F} = -2\hat{i} + 0\hat{j} = -2\hat{i} \text{ N}$.
- For $t > 2 \text{ s}: \vec{F} = 0\hat{i} + 0\hat{j} = 0 \text{ N}$.