Question

A runner taking part in the 200 m dash must run around the end of a track that has a circular arc with a radius of curvature of 50 m. If he completes the 200 m dash in 29.6 s and runs at constant speed throughout the race, what is the magnitude of his centripetal acceleration (in m/s2) as he runs the curved portion of the track?

Answer 1

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;

$a_c = \frac{V^2}{r}$

where;

r is the radius of the circular arc, given as 50 m

Substitute the givens;

$a_c = \frac{V^2}{r}\\\\a_c = \frac{(6.757)^2}{50}\\\\a_c = 0.91 \ m/s^2$

Therefore, the centripetal acceleration of the runner as he runs the curved portion of the track is 0.91 m/s².