Answer
A point on the edge of the wheel travels 33 meters.
Work Step by Step
$\Delta \omega = 360~rpm-240~rpm = 120~rpm$
$\Delta \omega = (120~rpm)(\frac{1~min}{60~s})$
$\Delta \omega = 2.0~rev/s$
We can find the angular acceleration:
$\alpha = \frac{\Delta \omega}{t}$
$\alpha = \frac{2.0~rev/s}{6.8~s}$
$\alpha = 0.294~rev/s^2$
We can convert $\omega_0 = 240~rpm$ to $rev/s$:
$\omega_0 = (240~rpm)(\frac{1~min}{60~s})$
$\omega_0 = 4.0~rev/s$
We can find the angular displacement:
$\theta = \omega_0 t + \frac{1}{2}\alpha t^2$ $\theta = (4.0~rev/s)(6.8~s) + \frac{1}{2}(0.294~rev/s^2)(6.8~s)^2$ $\theta = 34.0~rev$
We can find the distance $d$ traveled by a point on the edge of the wheel:
$d = \theta ~(2\pi~r)$
$d = (34.0~rev)(2)(\pi)(0.155~m)$
$d = 33~m$
A point on the edge of the wheel travels 33 meters.
