Chapter 8 linear appliCaTionS 293

Let t represent the number of hours after the second car has passed the
intersection.

65 13

13

65

1

5

t

t

t

=

=

=

In^15 of an hour or (^60) () 51  =  12 minutes, an additional 13 miles is between them.
Twelve minutes after the second car passes the intersection, there is a total of
18 miles between the cars. That is, at 6:22 the cars will be 18 miles apart.
Two employees ride their bikes to work. At 10:00 one leaves work and rides
southward home at 9 mph. At 10:05 the other leaves work and rides home
northward at 8 mph. When will they be 5 miles apart?
The first employee has ridden (^9) () 605  = ^34 miles by the time the second
employee has left. So we now need to see how long, after 10:05, it takes for
an additional 5  34 =  441  –  = ^174 miles to be between them. Let t represent
the number of hours after 10:05. When both employees are riding, the dis-
tance between them is increasing at the rate of 9 + 8 = 17 mph.
()98^17
4
17 17
4
1
17
17
4
1
4
+=

=⋅

t
t
t
t
After 41 hour, or 15 minutes, they will be an additional 441 miles apart. That
is, at 10:20, the employees will be 5 miles apart.
Two boys are 1250 meters apart when one begins walking toward the
other. If one walks at a rate of 2 meters per second and the other, who
starts walking toward the first boy four minutes later, walks at the rate of
1.5 meters per second, how long will it take for them to meet?
The boy with the head start has walked for 4(60) = 240 seconds. (Because
the rate is given in meters per second, we convert all times to seconds.) So,
he has traveled 240(2) = 480 meters. At the time the other boy begins walk-
ing, there remains 1250 – 480 = 770 meters to cover. When the second boy