A new car is tested on a 200 [m] diameter track. If the car speeds up at a steady 1.5 [m/s²], how long after starting is the magnitude of its centripetal acceleration equal to the tangential acceleration?
A new car is tested on a 200 [m] diameter track. If the car speeds up at a steady 1.5 [m/s²], how long after starting is the magnitude of its centripetal acceleration equal to the tangential acceleration?
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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![**Physics Problem: Centripetal and Tangential Acceleration**
**Question:**
A new car is tested on a 200-meter diameter track. If the car speeds up at a steady 1.5 meters per second squared, how long after starting is the magnitude of its centripetal acceleration equal to the tangential acceleration?
**Solution Approach:**
1. **Concepts to Understand:**
- **Tangential acceleration (at):** This is the rate at which the speed of the car increases and is given as 1.5 m/s².
- **Centripetal acceleration (ac):** This occurs when the car moves in a circular path and is directed towards the center of the circle. It can be calculated using the formula:
\[ ac = \frac{v^2}{r} \]
where \( v \) is the velocity and \( r \) is the radius of the circle.
2. **Given:**
- Diameter of the track = 200 m → Radius \( r = \frac{200}{2} = 100 \) m.
- Tangential acceleration \( at = 1.5 \, \text{m/s}^2 \).
3. **Find the time \( t \) when:**
- \( ac = at \)
4. **Equating Centripetal and Tangential Accelerations:**
\[ \frac{v^2}{r} = at \]
5. **Substituting known values:**
\[ \frac{v^2}{100} = 1.5 \]
6. **Solving for \( v \):**
\[ v^2 = 1.5 \times 100 \]
\[ v^2 = 150 \]
\[ v = \sqrt{150} \]
7. **Using the equation for velocity given constant acceleration:**
\[ v = at \]
\[ \sqrt{150} = 1.5 \times t \]
\[ t = \frac{\sqrt{150}}{1.5} \]
8. **Calculate \( t \).**
This approach gives a comprehensive method to determine the required time. Make use of a calculator to find the exact numerical solution.](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2Fa01c18cd-a3eb-4112-9efe-7c93aae07195%2F543335e1-35d4-4da1-bd5b-bad164cef7a6%2Fm05mj8a_processed.png&w=3840&q=75)
Transcribed Image Text:**Physics Problem: Centripetal and Tangential Acceleration**
**Question:**
A new car is tested on a 200-meter diameter track. If the car speeds up at a steady 1.5 meters per second squared, how long after starting is the magnitude of its centripetal acceleration equal to the tangential acceleration?
**Solution Approach:**
1. **Concepts to Understand:**
- **Tangential acceleration (at):** This is the rate at which the speed of the car increases and is given as 1.5 m/s².
- **Centripetal acceleration (ac):** This occurs when the car moves in a circular path and is directed towards the center of the circle. It can be calculated using the formula:
\[ ac = \frac{v^2}{r} \]
where \( v \) is the velocity and \( r \) is the radius of the circle.
2. **Given:**
- Diameter of the track = 200 m → Radius \( r = \frac{200}{2} = 100 \) m.
- Tangential acceleration \( at = 1.5 \, \text{m/s}^2 \).
3. **Find the time \( t \) when:**
- \( ac = at \)
4. **Equating Centripetal and Tangential Accelerations:**
\[ \frac{v^2}{r} = at \]
5. **Substituting known values:**
\[ \frac{v^2}{100} = 1.5 \]
6. **Solving for \( v \):**
\[ v^2 = 1.5 \times 100 \]
\[ v^2 = 150 \]
\[ v = \sqrt{150} \]
7. **Using the equation for velocity given constant acceleration:**
\[ v = at \]
\[ \sqrt{150} = 1.5 \times t \]
\[ t = \frac{\sqrt{150}}{1.5} \]
8. **Calculate \( t \).**
This approach gives a comprehensive method to determine the required time. Make use of a calculator to find the exact numerical solution.
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