The question asks for the number of times the minute-hand and the second-hand of a clock coincide between 09:15:00 AM and 09:45:00 AM.
First, determine the angular speed of each hand in degrees per second:
The minute and second hands cross when the faster second-hand overtakes the minute-hand. This requires the second-hand to gain a full 360 degrees relative to the minute-hand.
Calculate the relative speed:
Relative Speed = Speed(Second-hand) - Speed(Minute-hand)
Relative Speed = $6^\circ/\text{s} - 0.1^\circ/\text{s} = 5.9^\circ/\text{s}$.
The time between consecutive crossings is:
Time per Crossing = $\frac{360^\circ}{5.9^\circ/\text{s}} \approx 61.02$ seconds.
This implies they cross approximately once every minute. To be precise, within any 60-second interval [hh:mm:00, hh:mm+1:00), the second hand starts at 0 degrees and the minute hand is at some angle $\alpha$. They are guaranteed to cross exactly once within that interval.
The specified time interval is from 09:15:00 AM to 09:45:00 AM.
The total duration is 30 minutes.
The interval spans the following 1-minute periods where crossings occur:
The number of these 1-minute periods is $45 - 15 = 30$.
Since there is exactly one crossing per minute, there are 30 crossings in total.
The minute-hand and second-hand cross 30 times between 09:15:00 AM and 09:45:00 AM.