College Physics
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ISBN: 9781305952300
Author: Raymond A. Serway, Chris Vuille
Publisher: Cengage Learning
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![**Problem Statement:**
A car is traveling at a speed of 95.0 km/hr. The driver applies the brakes and the car comes to a stop over a distance of 60.0 meters.
**Question:**
What is the car's deceleration?
**Solution Explanation:**
To find the deceleration, we can use the formula:
\[ v^2 = u^2 + 2as \]
Where:
- \( v \) is the final velocity (0 m/s, since the car stops)
- \( u \) is the initial velocity (95.0 km/hr converted to m/s)
- \( a \) is the acceleration (deceleration value to be found)
- \( s \) is the stopping distance (60.0 m)
First, convert 95.0 km/hr to meters per second:
\[ 95.0 \, \text{km/hr} = \frac{95.0 \times 1000}{3600} \, \text{m/s} \]
Next, we rearrange the formula to solve for \( a \):
\[ 0 = u^2 + 2(-a)s \]
\[ a = \frac{-u^2}{2s} \]
Plug in the values to find the deceleration.](https://content.bartleby.com/qna-images/question/27e7e80b-03df-453b-a555-8f500263e07b/a9a0bf21-4972-4c0c-897f-879e8a42c233/hra9pr_processed.jpeg)
Transcribed Image Text:**Problem Statement:**
A car is traveling at a speed of 95.0 km/hr. The driver applies the brakes and the car comes to a stop over a distance of 60.0 meters.
**Question:**
What is the car's deceleration?
**Solution Explanation:**
To find the deceleration, we can use the formula:
\[ v^2 = u^2 + 2as \]
Where:
- \( v \) is the final velocity (0 m/s, since the car stops)
- \( u \) is the initial velocity (95.0 km/hr converted to m/s)
- \( a \) is the acceleration (deceleration value to be found)
- \( s \) is the stopping distance (60.0 m)
First, convert 95.0 km/hr to meters per second:
\[ 95.0 \, \text{km/hr} = \frac{95.0 \times 1000}{3600} \, \text{m/s} \]
Next, we rearrange the formula to solve for \( a \):
\[ 0 = u^2 + 2(-a)s \]
\[ a = \frac{-u^2}{2s} \]
Plug in the values to find the deceleration.
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