A car approaches the top of a hill that is shaped like a vertical circle with a radius of 59.5 m. What is the fastest speed that the car can go over the hill without losing contact with the ground? m/s
A car approaches the top of a hill that is shaped like a vertical circle with a radius of 59.5 m. What is the fastest speed that the car can go over the hill without losing contact with the ground? m/s
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 car approaches the top of a hill that is shaped like a vertical circle with a radius of 59.5 m. What is the fastest speed that the car can go over the hill without losing contact with the ground?*
___
**Solution:**
To solve this problem, we must determine the maximum speed at which a car can maintain contact with the ground at the top of a hill. This can be achieved by ensuring that the gravitational force is equal to the centripetal force required to keep the car on a circular path.
**Formula:**
The condition for the car to just stay in contact with the surface at the top of the hill is:
\[
mg = \frac{mv^2}{r}
\]
Where:
- \( m \) is the mass of the car,
- \( g \) is the acceleration due to gravity (\(9.8 \, \text{m/s}^2\)),
- \( v \) is the speed of the car,
- \( r \) is the radius of the circular hill.
**Solving for \( v \):**
\[
v = \sqrt{rg}
\]
**Given:**
\( r = 59.5 \, \text{m} \)
Substitute the values:
\[
v = \sqrt{59.5 \times 9.8} = \sqrt{583.1} \approx 24.14 \, \text{m/s}
\]
Therefore, the fastest speed the car can go over the hill without losing contact with the ground is approximately \(24.14 \, \text{m/s}\).](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2F8fd20e11-2399-44a5-beb6-e921116d74ba%2F6c3ddda3-6739-4a85-a8ed-ebab036a1507%2Fc6uhmn5_processed.jpeg&w=3840&q=75)
Transcribed Image Text:**Problem:**
*A car approaches the top of a hill that is shaped like a vertical circle with a radius of 59.5 m. What is the fastest speed that the car can go over the hill without losing contact with the ground?*
___
**Solution:**
To solve this problem, we must determine the maximum speed at which a car can maintain contact with the ground at the top of a hill. This can be achieved by ensuring that the gravitational force is equal to the centripetal force required to keep the car on a circular path.
**Formula:**
The condition for the car to just stay in contact with the surface at the top of the hill is:
\[
mg = \frac{mv^2}{r}
\]
Where:
- \( m \) is the mass of the car,
- \( g \) is the acceleration due to gravity (\(9.8 \, \text{m/s}^2\)),
- \( v \) is the speed of the car,
- \( r \) is the radius of the circular hill.
**Solving for \( v \):**
\[
v = \sqrt{rg}
\]
**Given:**
\( r = 59.5 \, \text{m} \)
Substitute the values:
\[
v = \sqrt{59.5 \times 9.8} = \sqrt{583.1} \approx 24.14 \, \text{m/s}
\]
Therefore, the fastest speed the car can go over the hill without losing contact with the ground is approximately \(24.14 \, \text{m/s}\).
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