Final answer:
Anne will catch up to Jay at 7:30 PM. This is found by calculating Jay's one-hour head start distance at 65 miles and determining that Anne, traveling at 75 mph, has a relative speed advantage of 10 mph. Anne needs 6.5 hours to catch up, so since she leaves at 1:00 PM, she will catch up to Jay at 7:30 PM, which is not listed among the provided options.
Step-by-step explanation:
To determine the time at which Anne will catch up to Jay, we need to calculate when their distances travelled will be equal, given that Jay has a 1-hour head start.
Jay’s head start can be calculated as the distance she travels in the first hour at 65 mph, which is:
d = r × t = 65 mph × 1 hour = 65 miles
This is the distance Anne needs to cover to catch up to Jay, in addition to the distance she covers while Jay travels at the same time. As Anne travels at 75 mph, her relative speed compared to Jay is:
Relative speed = Anne's speed - Jay's speed = 75 mph - 65 mph = 10 mph
The time it takes Anne to catch up is the head start distance divided by the relative speed:
Time to catch up = Distance to catch up / Relative speed = 65 miles / 10 mph = 6.5 hours
Since Anne leaves at 1:00 PM, she will catch up to Jay at:
1:00 PM + 6.5 hours = 7:30 PM
owever, since all the answer choices are in the afternoon, we need to verify our calculation since a 6.5-hour difference would mean she would catch up in the evening, which is not one of the options. The mistake here is that the time to catch up should not be in hours but fraction of an hour which is 6.5 hours equal to 6 hours and 30 minutes so:
1:00 PM + 6 hours and 30 minutes = 7:30 PM
The correct answer is not listed among the provided options.