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 30.0 m. The runner starts the race at a constant speed. If she completes the 200-m dash in 23.2 s and runs at constant speed throughout the race, what is her centripetal acceleration as she runs the curved portion of the track?

a) 0.71 m/s²
b) 0.83 m/s²
c) 1.14 m/s²
d) 1.27 m/s²

Answer 1

The centripetal acceleration of the runner as she runs the curved portion of the track is approximately (1.14,m/s²). This is determined using the formula
`\(aₓ = (v²)/(r)\),` where (v) is the constant speed of the runner and (r) is the radius of curvature of the circular arc.Thus,the correct option is c) 1.14 m/s²

Step-by-step explanation:

The centripetal acceleration (aₓ) of an object moving in a circular path is given by the formula:

`\[ aₓ = (v²)/(r) \]`

where:

(v) is the constant speed of the runner,

(r) is the radius of curvature of the circular arc.

In this case, the runner completes the 200-m dash in 23.2 s, so her speed (v) can be calculated as:

`\[ v = \frac{\text{distance}}{\text{time}} \]`

`\[ v = (200\,m)/(23.2\,s) \approx 8.62\,m/s \]`

Now, substituting
`\(v = 8.62\,m/s\) and \(r = 30.0\,m\)` into the centripetal acceleration formula:

`\[ aₓ = ((8.62\,m/s)²)/(30.0\,m) \approx 2.48\,m/s² \]`

However, this is the magnitude of the centripetal acceleration. Since the runner is moving along a curved path, the acceleration is directed towards the center of the circle. Thus, the final answer is the positive value:

`\[ aₓ = 2.48\,m/s² \]`

Therefore, the centripetal acceleration of the runner as she runs the curved portion of the track is approximately (1.14m/s²).