To determine the condition for collision avoidance, we analyze the motion in a reference frame moving with the truck. This simplifies the problem to a car decelerating towards a stationary target.

1. Determine the Car's Deceleration The car, with an initial speed of $v_c = 3v_0$, comes to a complete stop ($v_f = 0$) in a distance $s_1$. We use the time-independent kinematic equation, $v_f^2 = v_i^2 + 2ad$, to find the car's constant deceleration, $a_c$.

$$0^2 = (3v_0)^2 + 2a_c s_1$$

Solving for $a_c$:

$$a_c = -\frac{9v_0^2}{2s_1}$$

2. Analyze Relative Motion In the reference frame of the truck:

• The truck is stationary ($v_t' = 0$).
• The car's initial velocity relative to the truck is $v_{rel, i} = v_c - v_t = 3v_0 - v_0 = 2v_0$.
• The car's acceleration relative to the truck is $a_{rel} = a_c - a_t = a_c - 0 = a_c = -\frac{9v_0^2}{2s_1}$.

3. Find the Minimum Safe Distance A collision is avoided if the car can reduce its relative velocity to zero before or at the moment it covers the initial separation distance $s_0$. Let $s_{rel}$ be the distance the car travels relative to the truck to achieve a final relative velocity of zero ($v_{rel, f} = 0$). Using the same kinematic equation for relative motion:

$$v_{rel, f}^2 = v_{rel, i}^2 + 2a_{rel} s_{rel}$$ $$0^2 = (2v_0)^2 + 2\left(-\frac{9v_0^2}{2s_1}\right)s_{rel}$$ $$0 = 4v_0^2 - \frac{9v_0^2}{s_1}s_{rel}$$

Solving for the relative distance $s_{rel}$:

$$s_{rel} = \frac{4v_0^2 s_1}{9v_0^2} = \frac{4s_1}{9}$$

This is the minimum distance required for the car to match the truck's speed without collision.

4. State the Collision Avoidance Condition For the vehicles not to collide, the initial separation $s_0$ must be greater than or equal to this minimum required distance $s_{rel}$.

$$s_0 \ge s_{rel}$$

Therefore, the condition for collision avoidance is:

$$s_0 \ge \frac{4s_1}{9}$$