Question

3. A semicircular turn on a racetrack is 915 m long, and has a radius of 291 m. What is thecentripetal acceleration of a car that completes the turn at a constant speed in 15 s?

Answer 1

We are given that a car travels in circular motion is a semi-circle of radius 291 m and longitude of 915 m. To determine the centripetal acceleration we will use the following formula for the centripetal acceleration in a circular motion:

`a=(v^2)/(r)`

Where:

`\begin{gathered} a=\text{ acceleration} \\ v=\text{ velocity} \\ r=\text{ radius} \end{gathered}`

To determine the velocity we will use the fact that the car travels at a constant speed, therefore, the velocity is the quotient of the longitude of the track and the time it takes the car to travel through it:

`v=(s)/(t)`

Where:

`\begin{gathered} s=\text{ longitude} \\ t=\text{ time} \end{gathered}`

Now we plug in the values:

`v=(915m)/(15s)`

Solving the operations we get:

`v=61(m)/(s)`

Now we use this value in the formula for the acceleration:

`a=((61(m)/(s))^2)/(291m)`

Now we solve the operations and we get:

`a=12.79(m)/(s^2)`

Therefore, the centripetal acceleration of the car is 12.79 meters per second squared.