Question

A car travels at a constant speed around a circular track whose radius is 2.65 km. The car goes once around the track in 341 s. What is the magnitude of the centripetal acceleration of the car?

Answer 1

Concepts and reason

The concept of centripetal acceleration is used here. First, calculate the speed of the car by using the given radius and time. Then calculate the centripetal acceleration using the expression of centripetal acceleration.

Fundamentals

Centripetal acceleration:

The linear acceleration is the rate of change of linear velocity but the rate of change of tangential velocity is the centripetal acceleration. The centripetal makes an object to move in a curved path.

The expression for centripetal acceleration is:

${a_{\rm{c}}} = \frac{{{v^2}}}{r}$

Here, ${a_{\rm{c}}}$ is the centripetal acceleration, $v$ is the velocity and $r$ is the radial distance.

Total distance covered by the car is the circumference of the circular track.

$C = 2\pi r$

Substitute $2.65{\rm{ km}}$ for $r$ .

$\begin{array}{c}\\C = 2\pi \left( {2.65{\rm{ km}}} \right)\\\\ = 2\pi \left( {2.65 \times {{10}^3}{\rm{ m}}} \right)\\\\ = 16650.44{\rm{ m}}\\\end{array}$

Speed of the car is total distance divided by total time.

$v = \frac{C}{t}$

Substitute $16650.44{\rm{ m}}$ $C$ $341{\rm{ s}}$ for $t$ .

$\begin{array}{c}\\v = \frac{{16650.44{\rm{ m}}}}{{341{\rm{ s}}}}\\\\ = 48.828{\rm{ m/s}}\\\end{array}$

The centripetal acceleration is:

${a_{\rm{c}}} = \frac{{{v^2}}}{r}$

Substitute $48.828{\rm{ m/s}}$ for $v$ and $2.65 \times {10^3}{\rm{ m}}$ for $r$ .

$\begin{array}{c}\\{a_{\rm{c}}} = \frac{{{{\left( {48.828{\rm{ m/s}}} \right)}^2}}}{{2.65 \times {{10}^3}{\rm{ m}}}}\\\\ = \frac{{2384.173{\rm{ }}{{\rm{m}}^2}{\rm{/}}{{\rm{s}}^{\rm{2}}}}}{{2.65 \times {{10}^3}{\rm{ m}}}}\\\\ = 0.899{\rm{ m/}}{{\rm{s}}^2}\\\\ \simeq {\bf{0}}{\bf{.90 m/}}{{\bf{s}}^{\bf{2}}}\\\end{array}$

Ans:

The centripetal acceleration is ${\bf{0}}{\bf{.90 m/}}{{\bf{s}}^{\bf{2}}}$ .