A red car moving at 57.7 mi/hr passes a blue car moving at 37.8 mi/hr. (Use 1609 m = 1 mile and 1 hr = 3600 s when converting to meters/second.) Seeing a stopped car in his lane up ahead, the driver of the red car applies the brakes just as he passes the blue car which causes the red car to slow down to a stop by losing 4.24 m/s every second. How much time (in seconds) passes before the two cars are side-by-side once again?

Time

s

To solve this problem, we need to determine how much time will pass before the two cars are side-by-side again after the red car applies its brakes. 

### Step 1: Convert velocities to meters per second (m/s)
1. Convert the speed of the red car:
   \[
   \text{Speed of red car} = 57.7 \, \text{mi/hr} \times \left(\frac{1609 \, \text{m}}{1 \, \text{mi}}\right) \times \left(\frac{1 \, \text{hr}}{3600 \, \text{s}}\right)
   \]

2. Convert the speed of the blue car:
   \[
   \text{Speed of blue car} = 37.8 \, \text{mi/hr} \times \left(\frac{1609 \, \text{m}}{1 \, \text{mi}}\right) \times \left(\frac{1 \, \text{hr}}{3600 \, \text{s}}\right)
   \]

Let's calculate these conversions.The converted speeds are:
- Speed of the red car: \( 25.79 \, \text{m/s} \)
- Speed of the blue car: \( 16.89 \, \text{m/s} \)

### Step 2: Analyze the motion after braking
The red car starts decelerating with a constant rate of \( 4.24 \, \text{m/s}^2 \). The goal is to find the time when the red car and blue car are side by side again.

### Step 3: Set up equations of motion
1. **Red Car**: The red car decelerates from \( 25.79 \, \text{m/s} \) to a complete stop over time. We use the equation for motion under constant deceleration:
   \[
   v(t) = v_0 - a \cdot t
   \]
   where:
   - \( v_0 = 25.79 \, \text{m/s} \) (initial speed)
   - \( a = 4.24 \, \text{m/s}^2 \) (deceleration)
   - \( v(t) = 0 \) when the red car stops.

2. **Blue Car**: The blue car continues moving at a constant speed of \( 16.89 \, \text{m/s} \).

We need to determine when the two cars will be at the same position. Let's calculate the time it will take for the red car to stop and compare the positions of both cars over time to find when they meet.The time it takes for the red car to come to a complete stop is approximately \( 6.08 \, \text{seconds} \). However, the two cars will be side by side again after approximately \( 4.20 \, \text{seconds} \).

This means the red car will still be decelerating when they are side by side again, and it will take \( 4.20 \, \text{seconds} \) for the blue car to catch up.

Would you like further details or have any questions? Here are 5 related questions for you to consider:

1. How far did the red car travel before it came to a stop?
2. What distance did the blue car cover during the same time period?
3. How would the situation change if the deceleration rate of the red car was lower?
4. Could you calculate the stopping distance for the red car?
5. What would happen if the initial speed difference between the two cars was smaller?

**Tip:** When solving problems involving motion, separating the stages of motion (such as acceleration, deceleration, or constant velocity) can help clarify the steps needed for the solution.

A red car moving at 57.7 mi/hr passes a blue car moving at 37.8 mi/hr. (Use 1609 m = 1 mile and 1 hr = 3600 s when converting to meters/second.) Seeing a stopped car in his lane up ahead, the driver of the red car applies the brakes just as he passes the blue car which causes the red car to slow down to a stop by losing 4.24 m/s every second. How much time (in seconds) passes before the two cars are side-by-side once again?

Solve a detailed kinematics problem involving a red car decelerating from 57.7 mi/hr while a blue car moves at 37.8 mi/hr. This step-by-step solution covers motion under constant deceleration, conversion to meters per second, and calculating the time before the cars are side-by-side again. Ideal for advanced high school physics students.

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