Question

An object moving at a constant speed requires 4.0 s to go once around a circle with a diameter of 5.0 m. What is the magnitude of the instantaneous acceleration of the particle during this time? O2.20 m/s2 3.93 m/s2 6.17 m/s2 12.3m/2 15.4 m/g2

Answer 1

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 :

$a=r\omega^2$

Where

$\omega$ = angular velocity

$\omega=\dfrac{2\pi}{T}$

$a=\dfrac{4\pi^2r}{T^2}$

$a=\dfrac{4\pi^2\times 2.5\ m}{(4\ m)^2}$

$a=6.168\ m/s^2$

or

$a=6.17\ m/s^2$

So, the  instantaneous acceleration of the particle during this time is 6.17 m/s². Hence, this is the required solution.