To find the final separation distance between the two cars after both have stopped, we need to consider the movement of each car during the deceleration phase. This involves calculating the stopping distance for both cars, accounting for the green car’s reaction time delay, and the initial separation distance between them.

Step-by-step solution:

1. Convert all quantities to consistent units:

• The green car's speed is given in mi/hr, and we need to convert this to m/s. $v_{\text{green}} = 83.0 \, \text{mi/hr} = \frac{83.0}{2.24} \, \text{m/s} \approx 37.05 \, \text{m/s}$
• The yellow car's initial speed is the same as the green car’s, so: $v_{\text{yellow}} = 37.05 \, \text{m/s}$
• Convert the initial separation distance between the two cars from feet to meters: $d_{\text{initial}} = \frac{43.7 \, \text{ft}}{3.28} \approx 13.32 \, \text{m}$

2. Calculate the stopping distance for the yellow car:

The yellow car decelerates at $-9.08 \, \text{m/s}^2$. We can use the following kinematic equation to find the stopping distance: $v_f^2 = v_i^2 + 2a d$ Where:

• $v_f = 0$ m/s (final speed)
• $v_i = 37.05$ m/s (initial speed)
• $a = -9.08$ m/s² (deceleration)
• $d_{\text{yellow}} = \text{stopping distance of the yellow car}$

Rearranging for $d_{\text{yellow}}$: $d_{\text{yellow}} = \frac{v_i^2 - v_f^2}{2a} = \frac{(37.05)^2}{2 \times 9.08} \approx 75.63 \, \text{m}$

3. Calculate the stopping distance for the green car (after reaction delay):

The green car starts decelerating after a reaction time of $0.629$ seconds. During this time, the green car is still moving at $37.05 \, \text{m/s}$, so it covers some distance before it begins to decelerate.

The distance the green car travels during the reaction time is: $d_{\text{reaction}} = v_{\text{green}} \times t_{\text{reaction}} = 37.05 \times 0.629 \approx 23.31 \, \text{m}$

Now, we calculate the stopping distance for the green car after it begins decelerating at $-8.49 \, \text{m/s}^2$. Using the same kinematic equation: $d_{\text{green}} = \frac{(37.05)^2}{2 \times 8.49} \approx 80.91 \, \text{m}$

So, the total distance traveled by the green car is: $d_{\text{total, green}} = d_{\text{reaction}} + d_{\text{green}} = 23.31 + 80.91 \approx 104.22 \, \text{m}$

4. Calculate the final separation distance:

The initial separation between the cars is $13.32 \, \text{m}$. The green car travels a total of $104.22 \, \text{m}$, while the yellow car travels $75.63 \, \text{m}$.

Thus, the final separation distance is: $d_{\text{final}} = d_{\text{initial}} + (d_{\text{total, green}} - d_{\text{yellow}}) = 13.32 + (104.22 - 75.63) \approx 41.91 \, \text{m}$

Final Answer:

The final separation distance between the cars is approximately 41.91 meters.

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Would you like a detailed breakdown of any of these steps?

Here are 5 related questions to expand your knowledge:

1. How do reaction time and braking distance affect overall stopping distance?
2. What role does deceleration rate play in stopping time?
3. How does increasing the initial separation distance affect the final distance in this scenario?
4. What is the relationship between speed and stopping distance?
5. How would a different reaction time for the green car affect the result?

Tip: When calculating stopping distances, always ensure that all units (speed, time, distance) are consistent for accurate results.