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. The runner starts the race at a constant speed. If she completes the 200 m dash in 23.4 s and runs at constant speed throughout the race, what is the magnitude of her centripetal acceleration (in m/s2) as she runs the curved portion of the track

Answer 1

Answer:

$1.46m/s^{2}$

Explanation:

Data

Time:  $t=23.4s$<u />

Length:  $l=200m$

Radius: $r=50m$

We need to find centripetal acceleration

Velocity is the change of distance with time according to motion equation.

therefore

velocity:

$v=\frac{L}{t} \\v=\frac{200}{23.4}\\ v=8.55m/s$

The centripetal acceleration ($a_{c}$) is equal to the square of the velocity, divided by the radius of the circular path.

$a_{c} =\frac{v^{2} }{r} \\a_{a} =\frac{(8.55m/s)^{2} }{50m} \\a_{c} =\frac{73.10m^{2} s^{2} }{50m} \\a_{} =1.46m/s^{2}$

Answer 2

Answer:

Centripetal acceleration = 1.46m/s²

Explanation:

Given r = radius = 50m

v = distance/time = 200m/23.4s = 8.55m/s

Acceleration = v²/r = 8.55²/50 = 1.46m/s²