A car is going around a circular track at a constant speed of 17.5 m/s. The track has a radius of 75.5 meters. In units of meters per second squared, what is the magnitude of the acceleration of the car?
A car is going around a circular track at a constant speed of 17.5 m/s. The track has a radius of 75.5 meters. In units of meters per second squared, what is the magnitude of the acceleration of the car?
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![**Question:**
A car is going around a circular track at a constant speed of 17.5 m/s. The track has a radius of 75.5 meters. In units of meters per second squared, what is the magnitude of the acceleration of the car?
**Diagram Explanation:**
The diagram shows a circular track with a radius of 75.5 meters. A car is positioned on the circle's path, moving in a direction tangential to the circle. The velocity vector \( \vec{v} \) is depicted as an arrow pointing along the tangent to the circle, indicating the car's direction of motion relative to the circular path.
The diagram also includes a coordinate system with \( x \) and \( y \) axes intersecting at the center of the circle. The \( y \) axis is vertical, and the \( x \) axis is horizontal, creating a standard Cartesian plane centered on the circle's midpoint.
To solve for the magnitude of the acceleration:
The formula for centripetal acceleration (\( a_c \)) is:
\[
a_c = \frac{v^2}{r}
\]
Where:
- \( v \) = velocity of the car = 17.5 m/s
- \( r \) = radius of the circle = 75.5 m
Plug these values into the formula:
\[
a_c = \frac{(17.5)^2}{75.5} = \frac{306.25}{75.5} \approx 4.06 \, \text{m/s}^2
\]
Thus, the magnitude of the acceleration of the car is approximately \( 4.06 \, \text{m/s}^2 \).](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2F5c647828-f044-4f1b-b879-19c790819471%2F078e44f5-ac1c-4e06-9aec-f1a6c1f9dea6%2F21bisvd_processed.jpeg&w=3840&q=75)
Transcribed Image Text:**Question:**
A car is going around a circular track at a constant speed of 17.5 m/s. The track has a radius of 75.5 meters. In units of meters per second squared, what is the magnitude of the acceleration of the car?
**Diagram Explanation:**
The diagram shows a circular track with a radius of 75.5 meters. A car is positioned on the circle's path, moving in a direction tangential to the circle. The velocity vector \( \vec{v} \) is depicted as an arrow pointing along the tangent to the circle, indicating the car's direction of motion relative to the circular path.
The diagram also includes a coordinate system with \( x \) and \( y \) axes intersecting at the center of the circle. The \( y \) axis is vertical, and the \( x \) axis is horizontal, creating a standard Cartesian plane centered on the circle's midpoint.
To solve for the magnitude of the acceleration:
The formula for centripetal acceleration (\( a_c \)) is:
\[
a_c = \frac{v^2}{r}
\]
Where:
- \( v \) = velocity of the car = 17.5 m/s
- \( r \) = radius of the circle = 75.5 m
Plug these values into the formula:
\[
a_c = \frac{(17.5)^2}{75.5} = \frac{306.25}{75.5} \approx 4.06 \, \text{m/s}^2
\]
Thus, the magnitude of the acceleration of the car is approximately \( 4.06 \, \text{m/s}^2 \).
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