The time is 5:20. The acute angle formed by the clock hands is 40?. In exactly how many minutes will the big hand be right on top of the little hand? Show your work. Express your answer as a mixed number or as a decimal rounded to the nearest hundredth. Let N stand for “New angle”, let S stand for “Starting angle”, let m stand for time in minutes.
SOLUTION: When the big hand is right on top of the small hand with no space between them, they create an angle of 0?. So N, the new angle (i.e. the final angle) will be 0.?The time that will elapse in minutes is unknown. The starting angle is 40. The speeds of the two hands (see Problem 1) are: big hand, 6?/minute; small hand, .5?/minute. The big hand in this case is shrinking the angle, so we’ll subtract its movement from 40. The small hand will slightly grow the angle, so we’ll add its movement to 40. The general layout of the equation will be:
New angle = Starting angle – (big-hand-movement) + (small-hand-movement)
Plugging in our values:
0 = 40 – 6m + .5m . . . From here it’s just equation solving.
0 = 40 – 5.5m
5.5m = 40
m = 7 3/11 or 7.27 minutes