Chapter 5, Problem 1P

To determine

The magnitude of the centripetal acceleration of the car.

Answer to Problem 1P

  $0.8{\text{m/s}}^2$

Explanation of Solution

Given:

The radius of the circular track, $r=2.6\text{km}$

The time taken by the car to cover the track once, $T=360\text{s}$

Formula used:

The centripetal acceleration is given by,

  $a=\frac{v^2}{r}$

Where v is the speed of the object, and r is the radius of the curve covered by the object.

And, the time period of rotation on a circular track can be written in terms of linear speed as,

  $v=\frac{2\pi r}{T}$

Calculation:

Combining above two formulas, we get,

  $\begin{array}{l}a=\frac{{\left(2\pi r\right)}^2}{r{T}^2}\\ a=\frac{r{\left(2\pi \right)}^2}{T^2}\end{array}$

Using given values in above equation,

  $\begin{array}{l}a=\frac{2.6\times {10}^3{\left(2\pi \right)}^2}{{360}^2}\\ a=0.8{\text{m/s}}^2\end{array}$

Conclusion:

The magnitude of the centripetal acceleration of the car is $0.8{\text{m/s}}^2$ .