
A red car moving at 57.2 mi/hr passes a blue car moving at 38.7 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 4.72 m/s every second. How much time (in seconds) passes before the two cars are side-by-side once again?

We can break down the problem into two parts: converting the velocities of both cars into meters per second and then determining when they will be side-by-side again.

### Step 1: Convert velocities to meters per second

The velocities of the red and blue cars are given in miles per hour, so we first convert these into meters per second using the given conversion factors (1 mile = 1609 meters and 1 hour = 3600 seconds):

- Red car velocity: \( 57.2 \, \text{mi/hr} \)
  \[
  v_{\text{red}} = 57.2 \times \frac{1609 \, \text{m}}{1 \, \text{mi}} \times \frac{1 \, \text{hr}}{3600 \, \text{s}} = 25.57 \, \text{m/s}
  \]

- Blue car velocity: \( 38.7 \, \text{mi/hr} \)
  \[
  v_{\text{blue}} = 38.7 \times \frac{1609 \, \text{m}}{1 \, \text{mi}} \times \frac{1 \, \text{hr}}{3600 \, \text{s}} = 17.30 \, \text{m/s}
  \]

### Step 2: Analyze the situation after the red car applies the brakes

When the red car applies the brakes, it decelerates at \( 4.72 \, \text{m/s}^2 \) until it comes to a stop. We need to calculate how long it takes for the red car to decelerate to zero speed.

The red car's deceleration can be modeled by the equation:
\[
v_{\text{red final}} = v_{\text{red initial}} - a \cdot t
\]
where:
- \( v_{\text{red final}} = 0 \, \text{m/s} \) (final velocity of the red car),
- \( v_{\text{red initial}} = 25.57 \, \text{m/s} \) (initial velocity of the red car),
- \( a = 4.72 \, \text{m/s}^2 \) (deceleration).

Solving for \( t \) (time taken for the red car to stop):
\[
0 = 25.57 - 4.72 \cdot t
\]
\[
t = \frac{25.57}{4.72} = 5.42 \, \text{seconds}
\]

### Step 3: Determine when the two cars are side-by-side again

During this time, the blue car continues to travel at its constant speed. The red car is decelerating and will eventually stop.

The distance covered by the blue car during the time it takes for the red car to stop is:
\[
d_{\text{blue}} = v_{\text{blue}} \cdot t = 17.30 \, \text{m/s} \times 5.42 \, \text{s} = 93.85 \, \text{meters}
\]

The red car is decelerating, so the distance it travels before stopping can be calculated using the equation for distance under uniform deceleration:
\[
d_{\text{red}} = v_{\text{red initial}} \cdot t - \frac{1}{2} a \cdot t^2
\]
\[
d_{\text{red}} = 25.57 \cdot 5.42 - \frac{1}{2} \cdot 4.72 \cdot (5.42)^2 = 138.56 - 69.26 = 69.30 \, \text{meters}
\]

Since the red car only travels 69.30 meters before stopping, while the blue car covers 93.85 meters, the two cars will be side-by-side once again after 5.42 seconds, when the red car stops.

### Final Answer:
The time that passes before the two cars are side-by-side once again is **5.42 seconds**.

Would you like more details or have any questions?

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Here are 5 relative questions for practice:

1. How far would the blue car have traveled after 10 seconds at a constant speed?
2. What would happen if the red car’s deceleration rate was increased to 6 m/s²?
3. How long would it take for the red car to stop if it was initially traveling at 60 mi/hr?
4. What is the total distance covered by both cars after 8 seconds?
5. How would the result change if the red car slowed down by 3 m/s² instead of 4.72 m/s²?

**Tip**: When solving motion problems, it's often helpful to convert all units to SI (meters and seconds) before starting.

A red car moving at 57.2 mi/hr passes a blue car moving at 38.7 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 4.72 m/s every second. How much time (in seconds) passes before the two cars are side-by-side once again?

Solve a kinematics problem involving a red car decelerating from 57.2 mi/hr to a stop, while a blue car continues at 38.7 mi/hr. Learn how to calculate the time it takes for the two cars to be side-by-side again, using velocity conversion, deceleration equations, and uniform motion principles.

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