Question

john and karen are traveling south in separate cars. karent travels at 65 mph and john 70mph. karen passes exit 3 at 1:15 and john passes same exit at 1:30. what time will john catch up with karen?

Answer 1

Let's set 1:15 as the zero point for a variable t representing time. So, Karen passed the exit at t=0, and John passed it at t=.25 (in hours). Karen's distance from the exit can be represented as 65t, and John's can be represented as 70(t-.25). Set the two expressions equal to each other to find out when they intersect. 65t=70(t-.25), so 65t=70t-17.5, 5t=17.5, t=3.5. So, John will catch up with Karen 3.5 hours after 1:15, which is 4:45.

Alternative intuition: When John passes the exit, Karen has gone 65 mph for a quarter hour, giving her a 16.25 mile head start. John is traveling 5 mph faster than Karen. At 5 mph, that 16.25 mile head start will last 16.25/5 hours, or 3.25 hours after John passes the exit. 3.25 hours after 1:30, which is also 4:45.