We are tasked with determining the initial velocity $v_{g0}$ and the constant acceleration $a_g$ of the green car, given the motion of both the red and green cars along the x-axis.

Given Information:

1. Initial position of the red car $x_r(0) = 0$ m.
2. Initial position of the green car $x_g(0) = 218$ m.
3. Case 1: Red car's velocity $v_r = 27.0 \text{ km/h} = 7.5 \text{ m/s}$, they pass at $x = 43.0$ m.
4. Case 2: Red car's velocity $v_r = 54.0 \text{ km/h} = 15.0 \text{ m/s}$, they pass at $x = 76.7$ m.
5. The green car has a constant acceleration $a_g$, but we don't know its value yet.

Setup the equations of motion

1. Red car's equation of motion

The red car moves with constant velocity, so its position at any time $t$ is given by: $x_r(t) = v_r t$ where $v_r$ is the constant velocity of the red car.

2. Green car's equation of motion

The green car starts at $x_g(0) = 218$ m with an initial velocity $v_{g0}$ and a constant acceleration $a_g$. The equation for its motion is: $x_g(t) = 218 + v_{g0} t + \frac{1}{2} a_g t^2$

Condition for passing each other

The two cars pass each other when their positions are equal. That means: $x_r(t) = x_g(t)$ For both scenarios, we will set up this equality at the points they pass each other.

Case 1: $v_r = 7.5$ m/s, they pass at $x = 43.0$ m.

At the moment they pass, the red car is at $x = 43.0$ m. So: $x_r(t) = 43.0 = 7.5 t_1 \quad \Rightarrow \quad t_1 = \frac{43.0}{7.5} = 5.73 \, \text{s}$ At this time $t_1 = 5.73$ s, the green car is also at $x = 43.0$ m: $43.0 = 218 + v_{g0} \cdot 5.73 + \frac{1}{2} a_g \cdot (5.73)^2$ This gives us our first equation.

Case 2: $v_r = 15.0$ m/s, they pass at $x = 76.7$ m.

At this moment, the red car is at $x = 76.7$ m. So: $x_r(t) = 76.7 = 15.0 t_2 \quad \Rightarrow \quad t_2 = \frac{76.7}{15.0} = 5.11 \, \text{s}$ At this time $t_2 = 5.11$ s, the green car is also at $x = 76.7$ m: $76.7 = 218 + v_{g0} \cdot 5.11 + \frac{1}{2} a_g \cdot (5.11)^2$ This gives us our second equation.

Solving the system of equations

We now have two equations with two unknowns, $v_{g0}$ and $a_g$:

1. $43.0 = 218 + 5.73 v_{g0} + \frac{1}{2} a_g \cdot (5.73)^2$
2. $76.7 = 218 + 5.11 v_{g0} + \frac{1}{2} a_g \cdot (5.11)^2$

Let me solve this system of equations for $v_{g0}$ and $a_g$.The solutions for the initial velocity and acceleration of the green car are:

• (a) The initial velocity of the green car: $v_{g0} = -3.84 \, \text{m/s}$
• (b) The constant acceleration of the green car: $a_g = -9.32 \, \text{m/s}^2$

The negative signs indicate that both the initial velocity and the acceleration of the green car are directed opposite to the positive x-axis.

Would you like more details or have any questions?

Here are 5 related questions to expand on this:

1. How does the negative acceleration of the green car affect its motion?
2. What would happen if both cars had the same acceleration?
3. How would the result change if the green car had a positive acceleration?
4. Can you derive the equation for the green car's displacement with different accelerations?
5. How can you interpret the physical meaning of the time intervals?

Tip: When solving kinematic problems, always pay attention to the signs of velocity and acceleration. They indicate the direction of motion and change.