Question

What is the magnitude of the acceleration of a sprinter running at \[10m \cdot {s^{ - 1}}\] when rounding a turn of radius \[25\,m\] ?

Answer 1

Hint: In order to answer this question to find the magnitude of the acceleration of the sprinter we will use the formula of centripetal acceleration. This sprinter's acceleration is the same as the centripetal acceleration. The uniform circular motion is the source of the formula. The centripetal acceleration is equal to the square of the body's speed divided by the radius of the circular track.

Complete step by step answer:
Remember that acceleration is defined as a change in velocity. The direction of your velocity changes when you round a corner, indicating that you are accelerating. There will be centripetal (towards the centre) acceleration since the sprinter is making a circular turn. This has the form:
${a_c} = \dfrac{{{v^2}}}{r} = m{\omega ^2}r$
Where, $v$ is the runner’s velocity, $r$ is the radius of the turn, and $\omega $ is the angular velocity of the turn.

So, now for the sprinter the magnitude of the acceleration is given by,
${a_c} = \dfrac{{{v^2}}}{r}$
where \[v\] is velocity and \[r\] is radius of the track.
${a_c} = \dfrac{{{v^2}}}{r} \\
\Rightarrow {a_c} = \dfrac{{\left( {10m \cdot {s^{ - 2}}} \right)}}{{\left( {25m} \right)}} \\
\therefore {a_c} = 4.0\,m \cdot {s^{ - 2}}$
Therefore, the magnitude of the acceleration of a sprinter running is $4.0m \cdot {s^{ - 2}}$.

Note: The term "acceleration" refers to a shift in velocity. It may occur to you to wonder how something travelling at a constant speed in a circle can have an acceleration. Speed is the rate at which an item moves and is scalar because it has no direction. Velocity is the speed as well as the direction, and it is a vector because it has no direction.