-vel 5. Ablue car and a red car are racing. They start in the same place.at the same time. The blue car starts from rest and the red car starts at 5 m/s. The blue car accelerates at a rate of 4 m/s/s, and the red car maintains its speed. When does the blue car catch up to the red car?

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**Problem:**

A blue car and a red car are racing. They start in the same place at the same time. The blue car starts from rest, and the red car starts at 5 m/s. The blue car accelerates at a rate of 4 m/s², and the red car maintains its speed. When does the blue car catch up to the red car?

**Given Information:**

- **Blue Car:**
  - Initial velocity = 0 m/s (starts from rest)
  - Acceleration = 4 m/s²

- **Red Car:**
  - Constant velocity = 5 m/s

**Solution:**

Let's set up equations for the distance traveled by each car and find when they are equal, which is when the blue car catches up.

**Distance Formula:**

- For the blue car with constant acceleration:
  \[ d_{\text{blue}} = v_i \cdot t + \frac{1}{2} a \cdot t^2 \]
  \[ d_{\text{blue}} = 0 \cdot t + \frac{1}{2} \cdot 4 \cdot t^2 \]
  \[ d_{\text{blue}} = 2t^2 \]

- For the red car with constant speed:
  \[ d_{\text{red}} = v \cdot t \]
  \[ d_{\text{red}} = 5t \]

**Set the distances equal to find when they are the same:**

\[ 2t^2 = 5t \]

Solving for \( t \):
\[ 2t^2 - 5t = 0 \]
\[ t(2t - 5) = 0 \]

This gives possible solutions for \( t \):
- \( t = 0 \) (when they start together, but not when the blue car catches up)
- \( 2t - 5 = 0 \Rightarrow t = 2.5 \)

Thus, the blue car catches up to the red car after 2.5 seconds.
Transcribed Image Text:**Problem:** A blue car and a red car are racing. They start in the same place at the same time. The blue car starts from rest, and the red car starts at 5 m/s. The blue car accelerates at a rate of 4 m/s², and the red car maintains its speed. When does the blue car catch up to the red car? **Given Information:** - **Blue Car:** - Initial velocity = 0 m/s (starts from rest) - Acceleration = 4 m/s² - **Red Car:** - Constant velocity = 5 m/s **Solution:** Let's set up equations for the distance traveled by each car and find when they are equal, which is when the blue car catches up. **Distance Formula:** - For the blue car with constant acceleration: \[ d_{\text{blue}} = v_i \cdot t + \frac{1}{2} a \cdot t^2 \] \[ d_{\text{blue}} = 0 \cdot t + \frac{1}{2} \cdot 4 \cdot t^2 \] \[ d_{\text{blue}} = 2t^2 \] - For the red car with constant speed: \[ d_{\text{red}} = v \cdot t \] \[ d_{\text{red}} = 5t \] **Set the distances equal to find when they are the same:** \[ 2t^2 = 5t \] Solving for \( t \): \[ 2t^2 - 5t = 0 \] \[ t(2t - 5) = 0 \] This gives possible solutions for \( t \): - \( t = 0 \) (when they start together, but not when the blue car catches up) - \( 2t - 5 = 0 \Rightarrow t = 2.5 \) Thus, the blue car catches up to the red car after 2.5 seconds.
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