Chapter 8 linear appliCaTionS 295✔SOLUTIONS
The first jogger had jogged (^7) () 608 = ^56601415 miles when the other jog-=
ger began. So there is 2 1415 = 115 – = 1516 miles left to cover. The distance
between them is growing at a rate of 7 + 9 = 16 mph. Let t represent the
number of hours the second jogger jogs.
16 16
15
1
16
16
15
1
15
t
t
t
=⋅
In 151 of an hour, or (^60) () 151 = 4 minutes after the second jogger began,
the joggers will be two miles apart.
The first boat got a (^12) () 604 = ^45 mile head start. When the second boat
leaves, there remains 6 45 = 551 – = ^265 miles between them. When the
second boat leaves the distance between them is decreasing at a rate
of 12 + 14 = 26 mph. Let t represent the number of hours the second
boat travels.
26 26
5
1
26
26
5
1
5
t
t
t
=⋅
In^15 hour, or 51 () 60 = 20 minutes, after the second boat leaves, the boats
will meet. That is, at 2:29 both boats will meet.
The Smiths have driven (^55) ()^1260 = 11 miles outside of Tulsa by the time
the Hewitts have left Dallas. So, when the Hewitts leave Dallas, there are
257 − 11 = 246 miles between the Smiths and Hewitts. When the Hewitts
leave Dallas, the distance between them is decreasing at the rate of
55 + 65 = 120 mph. Let t represent the number of hours after the Hewitts
have left Dallas.
120 246
246
120
2201
t
t
t
=