-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?
-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?
College Physics
11th Edition
ISBN:9781305952300
Author:Raymond A. Serway, Chris Vuille
Publisher:Raymond A. Serway, Chris Vuille
Chapter1: Units, Trigonometry. And Vectors
Section: Chapter Questions
Problem 1CQ: Estimate the order of magnitude of the length, in meters, of each of the following; (a) a mouse, (b)...
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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.](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2Fd219e951-5960-43a0-847a-c0317cc0d5af%2Fb84a03a6-6466-44d7-b6d0-08bac972ee47%2F1g6hbb.jpeg&w=3840&q=75)
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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