### Problem StatementA race car travels with a constant tangential speed of 73.7 m/s around a circular track of radius 646 m. Find the magnitude of the total acceleration.### Diagram DescriptionThe diagram illustrates a race car moving on a circular track. The track is represented as a circle with a radius labeled "r" and an internal circular path depicted. The tangential speed, denoted as \( v_T \), is represented by an arrow pointing along the tangent to the circle at the position of a small dot indicating the race car on the track. The radius "r" extends from the center of the circle to the position of the race car on the track, perpendicular to the tangential speed vector \( v_T \).### Label- \( v_T \): Tangential speed (73.7 m/s)- r: Radius of the circular track (646 m)### Input Fields- **Number**: Input field to enter the calculated magnitude of the total acceleration.- **Units**: Dropdown menu to select the units of the calculated acceleration (e.g., m/s²).### Solution MethodTo find the magnitude of the total acceleration of the race car, we need to consider its centripetal acceleration because the car is moving in a circular path with a constant tangential speed.The formula for centripetal acceleration \( a_c \) is:\[ a_c = \frac{v_T^2}{r} \]Where:- \( v_T \) is the tangential speed, 73.7 m/s.- \( r \) is the radius of the circular track, 646 m.Substitute the given values into the formula:\[ a_c = \frac{(73.7 \, \text{m/s})^2}{646 \, \text{m}} \]Calculating the value:\[ a_c = \frac{5431.69}{646} \, \text{m/s}^2 \approx 8.41 \, \text{m/s}^2 \]Since the tangential speed is constant, the tangential acceleration is zero. Therefore, the total acceleration of the race car is entirely due to the centripetal acceleration.### Final AnswerThe magnitude of the total acceleration is approximately 8.41 m/s².

Given that Constant tangential speed V = 73.7 m/s Radius of circular track R = 646 m

Answered: A race car travels with a constant…

A race car travels with a constant tangential speed of 73.7 m/s around a circular track of radius 646 m. Find the magnitude of tHe total 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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Concept explainers

Centripetal Force

A particle, body, object or certain mass while travelling in a rotational path or semicircular path, experiences an inertia force. If that inertia force pulls the towards the inward of the imaginary circle in the rotation, it is termed as centripetal force.

Question

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### Problem Statement

A race car travels with a constant tangential speed of 73.7 m/s around a circular track of radius 646 m. Find the magnitude of the total acceleration.

### Diagram Description

The diagram illustrates a race car moving on a circular track. The track is represented as a circle with a radius labeled "r" and an internal circular path depicted. The tangential speed, denoted as \( v_T \), is represented by an arrow pointing along the tangent to the circle at the position of a small dot indicating the race car on the track. The radius "r" extends from the center of the circle to the position of the race car on the track, perpendicular to the tangential speed vector \( v_T \).

### Label
- \( v_T \): Tangential speed (73.7 m/s)
- r: Radius of the circular track (646 m)

### Input Fields
- **Number**: Input field to enter the calculated magnitude of the total acceleration.
- **Units**: Dropdown menu to select the units of the calculated acceleration (e.g., m/s²).

### Solution Method

To find the magnitude of the total acceleration of the race car, we need to consider its centripetal acceleration because the car is moving in a circular path with a constant tangential speed.

The formula for centripetal acceleration \( a_c \) is:
\[ a_c = \frac{v_T^2}{r} \]

Where:
- \( v_T \) is the tangential speed, 73.7 m/s.
- \( r \) is the radius of the circular track, 646 m.

Substitute the given values into the formula:
\[ a_c = \frac{(73.7 \, \text{m/s})^2}{646 \, \text{m}} \]

Calculating the value:
\[ a_c = \frac{5431.69}{646} \, \text{m/s}^2 \approx 8.41 \, \text{m/s}^2 \]

Since the tangential speed is constant, the tangential acceleration is zero. Therefore, the total acceleration of the race car is entirely due to the centripetal acceleration.

### Final Answer
The magnitude of the total acceleration is approximately 8.41 m/s².

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