Total distance = \( s \cdot t \)
Initial speed is \( \frac{s}{3} \). Assuming time for this segment is 5 minutes (\( \frac{1}{12} \) hour).
Distance covered in this time: \( \frac{s}{3} \cdot \frac{1}{12} = \frac{s}{36} \)
Assuming standard time is 1 hour, remaining time is \( \frac{11}{12} \) hour.
To compensate for the delay, the required speed \( v \) must cover the remaining distance:
\[ v = \frac{\text{Remaining distance}}{\text{Time left}} = \frac{s \cdot t - \frac{s}{36}}{t - \frac{1}{12}} \]
Alternatively, using the proportion method:
The last \( \frac{2st}{3} \) distance must be covered in \( \frac{2t}{5} \) time (6 minutes).
Required speed: \[ = \frac{\frac{2st}{3}}{\frac{2t}{5}} = \frac{5s}{3} \]
\[ \text{Increase} = \frac{\frac{5s}{3} - s}{s} \cdot 100 = \frac{2s}{3s} \cdot 100 = \frac{2}{3} \cdot 100 = 66.\overline{6}\% \]
Correct Option: (B)