Answer
The correct answer is:
D. $50~m/s^2$
Work Step by Step
We can find the tangential speed $v$ of a point on the wheel at a distance of $0.3~m$ from the center:
$v = \frac{d}{t} = \frac{4800~m}{(20)(60~s)} = 4.0~m/s$
The can find the radial acceleration:
$a_r = \frac{v^2}{r} = \frac{(4.0~m/s)^2}{0.3~m} = 53.3~m/s^2$
This value is closest to $50~m/s^2$
The correct answer is:
D. $50~m/s^2$
