Answer:
The centripetal acceleration of the runner as he runs the curved portion of the track is 0.91 m/s²
Explanation:
Given;
distance traveled in the given time = 200 m
time to cover the distance, t = 29.6 s
speed of the runner, v = d / t
v = 200 / 29.6
v = 6.757 m/s
The centripetal acceleration of the runner is given by;
[tex]a_c = \frac{V^2}{r}[/tex]
where;
r is the radius of the circular arc, given as 50 m
Substitute the givens;
[tex]a_c = \frac{V^2}{r}\\\\a_c = \frac{(6.757)^2}{50}\\\\a_c = 0.91 \ m/s^2[/tex]
Therefore, the centripetal acceleration of the runner as he runs the curved portion of the track is 0.91 m/s².
