Answer:
Acceleration, a = 6.17 m/s²
Explanation:
It is given that,
Diameter of the circular track, d = 5 m
Radius of circular track, r = 2.5 m
An object moving at a constant speed requires 4.0 s to go once around a circle. We need to find the instantaneous acceleration of the particle during this time. It is given by :
[tex]a=r\omega^2[/tex]
Where
[tex]\omega[/tex] = angular velocity
[tex]\omega=\dfrac{2\pi}{T}[/tex]
[tex]a=\dfrac{4\pi^2r}{T^2}[/tex]
[tex]a=\dfrac{4\pi^2\times 2.5\ m}{(4\ m)^2}[/tex]
[tex]a=6.168\ m/s^2[/tex]
or
[tex]a=6.17\ m/s^2[/tex]
So, the instantaneous acceleration of the particle during this time is 6.17 m/s². Hence, this is the required solution.
