Let the motion of the particle be described by the equation of motion for constant velocity, where $x$ is position and $t$ is time:

We are given three conditions that constrain the possible values of the initial position $x_0$ and the velocity $v$.

• Condition I (Start): At $t=0$, the position is $10 \le x_0 \le 12$.
• Condition II (4th Second): The particle passes $x=22$ at some instant $t_1$ during the fourth second ($3 \le t_1 \le 4$). $$ x(t_1) = x_0 + v t_1 = 22 \implies x_0 = 22 – v t_1 $$ Since $3 \le t_1 \le 4$, we have the inequality: $$ 22 – 4v \le x_0 \le 22 – 3v $$
• Condition III ($t=12$): At $t=12$, the position is $55 \le x(12) \le 60$. $$ 55 \le x_0 + 12v \le 60 \implies 55 – 12v \le x_0 \le 60 – 12v $$

Combining the constraints allows us to find the range of valid velocities:

• Minimum Velocity ($v_{min}$): Connects the “latest” possible pass at $x=22$ ($t=3$) with the “lowest” possible point at $t=12$. $$ v \ge \frac{11}{3} \approx 3.67 \, \text{m/s} $$ Corresponding start position: $x_0 = 11$ m.
• Maximum Velocity ($v_{max}$): Constrained by the start point $x_0=10$ and passing $x=22$ as early as possible ($t=3$). $$ v \le 4 \, \text{m/s} $$ Corresponding start position: $x_0 = 10$ m.

Using $t = \frac{88 – x_0}{v}$, we find the arrival times:

Instead of full inequality modeling, we can geometrically identify the “fastest” and “slowest” possible lines by connecting the extreme boundary points of the constraints.