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1) Suppose that the spool is rotating with an angular speed \omega about a point A. The spool is a rigid body, this means that the angular speed is the same for all points. Write equations for points B and C:
\omega_C = \frac{v_C}{R}, \omega_B = \frac{v_B}{R + r}
but \omega_B = \omega_C = \omega and v_B = u because the board does not slip on the spool. Thus, v_C = \frac{R}{R+r}u. This means that the speed at which the man is approching the spool is v = u - v_C = u -\frac{R}{R+r}u =\frac{r}{R+r}u. This means that the time needed for the man to reach the spool is t = \frac{l}{v} = \frac{l(R+r)}{ru}\;.
2) The man will travel s = ut = \frac{l(R+r)}{r}.
3) If r = R then s = 2l.
4) Plug numbers to the equations but do not forget to change units, t = 6.48\mathrm{s} and s = 6.48\mathrm{m}.
Investigate the relationship between speeds recorded and the distance travelled in this kinematic scenario.