Starting at \(3\text{ o'clock}\), at what time will the minute hand catch up with the hour hand?

The tip of the minute hand travels in a circular path as it moves around the clock. If the "\(12\)" marker on the clock face is considered to be \(0^{\circ}\), then the "\(3\)" marker will be at \(90^{\circ}\), the "\(6\)" marker at \(180^{\circ}\), and the "\(9\)" marker at \(270^{\circ}\). When the minute hand overtakes the hour hand, both will be at the same degree marker.

Let \(P=\) the number of minutes the minute hand takes to catch up with the hour hand.

The minute hand covers \(360^{\circ}\) in \(60\text{ minutes}\), or \(360/60=6\text{ degrees per minute}\). So in \(P\text{ minutes}\) it will have covered \((P)(6)\) or \(6P\text{ degrees}\), and will be at the \(6P\) degree mark since it started at the \(0^{\circ}\) mark..

The hour hand covers \(360^{\circ}\) in \((60)(12)\) or \(720\text{ minutes}\), so is moving at a rate of \(360/720=0.5\text{ degrees per minute}\). So in \(P\text{ minutes}\), the hour hand will have moved \((P)(0.5)\) or \(0.5P\text{ degrees}\) past the "\(3\)" marker. Since the "\(3\)" marker is at the \(90^{\circ}\) point, the hour hand will be at the \((90 + 0.5P)\) degree mark after \(P\text{ minutes}\)..

When the minute hand catches up with the hour hand, both will be at the same degree point, therefore,

\begin{align*} 90+0.5P&=6P\\ \text{or, }P&=90/5.5\\ &=16.3636\text{ minutes} \end{align*}